Olympic-level physics question:
Q:-1. "In the sport of swimming, a swimmer is attempting to break the world record in the 100m freestyle event. If the swimmer starts from rest and accelerates uniformly, reaching a final speed of 10 m/s at the end of the race, calculate the average acceleration of the swimmer during the race. Assume the race is a straight line and there are no external forces affecting the swimmer’s motion."
To solve this problem, we'll use the kinematic equation for uniformly accelerated motion:
v_f = v_i + a * t
Given:
Initial speed, v_i = 0 m/s
Final speed, v_f = 10 m/s
Distance, d = 100 m
Step 1: Calculate the time taken to reach the final speed.
Using the kinematic equation above, we can rearrange the formula to solve for time (t):
t = (v_f - v_i) / a
Step 2: Calculate the average acceleration.
The average acceleration can be calculated as the change in velocity divided by the time taken:
a = (v_f - v_i) / t
Step 3: Solve for acceleration.
Substitute the given values into the equation:
a = (10 m/s - 0 m/s) / t
We need to find the time taken to cover a distance of 100m. The distance traveled can be calculated using the formula:
d = v_i * t + (1/2) * a * t^2
Since the initial speed is zero, the formula simplifies to:
d = (1/2) * a * t^2
Now, rearrange the formula to solve for time (t):
t = sqrt((2 * d) / a)
Substitute the known values into the equation:
t = sqrt((2 * 100 m) / a)
Now, substitute the value of t obtained above into the equation for acceleration:
a = (10 m/s - 0 m/s) / sqrt((2 * 100 m) / a)
Now, solve for acceleration.
Therefore, the average acceleration of the swimmer during the race can be determined by solving the equation (10 m/s - 0 m/s) / sqrt((2 * 100 m) / a), where v_f is the final speed, v_i is the initial speed, d is the distance, and a is the average acceleration.
Q:- 2. "In the sport of pole vaulting, an athlete uses a flexible carbon fiber pole to clear the bar. The athlete approaches the takeoff point with an initial speed of 10 m/s. If the pole bends to an angle of 45 degrees with respect to the ground during the vault, calculate the change in the athlete's gravitational potential energy when reaching the highest point of the vault. Assume the athlete's mass is 70 kg and the acceleration due to gravity is 9.8 m/s^2."
To solve this problem, we need to consider the change in gravitational potential energy as the athlete reaches the highest point of the pole vault.
Given:
Initial speed, v_i = 10 m/s
Pole bend angle, θ = 45 degrees
Mass of the athlete, m = 70 kg
Acceleration due to gravity, g = 9.8 m/s^2
Step 1: Determine the height reached by the athlete during the vault.
To calculate the height, we need to find the vertical component of the initial velocity. Using trigonometry:
v_iy = v_i * sin(θ)
Step 2: Calculate the final velocity at the highest point.
At the highest point of the vault, the vertical velocity component will be zero, while the horizontal component will remain the same as the initial velocity. Therefore, the final velocity (v_f) will have a magnitude equal to the initial horizontal velocity but in the opposite direction.
Step 3: Calculate the change in gravitational potential energy.
The change in gravitational potential energy can be given as:
ΔPE = m * g * Δh
where Δh is the change in height. Since the vault reaches the highest point, Δh is essentially the height reached during the vault.
Substituting the known values into the formula:
ΔPE = m * g * Δh
ΔPE = m * g * (2 * v_iy / g) [Using the kinematic equation: Δh = (v_f^2 - v_i^2) / (2 * g), where v_f = 0 at the highest point]
ΔPE = m * 2 * v_iy
ΔPE = 2 * m * v_i * sin(θ)
Now, substitute the values and calculate the change in gravitational potential energy:
ΔPE = 2 * 70 kg * 10 m/s * sin(45°)
Therefore, the change in the athlete's gravitational potential energy when reaching the highest point of the vault is approximately 980 J (joules).
Q:-3. "In the high jump event, an athlete approaches the bar with an initial speed of 9 m/s at an angle of 30 degrees above the horizontal. If the athlete clears the bar at a height of 2 meters, calculate the horizontal distance from the takeoff point to where the athlete clears the bar. Assume air resistance is negligible."
To solve this problem, we'll use the kinematic equations for projectile motion and apply them to the vertical and horizontal components separately.
Given:
Initial speed, v_i = 9 m/s
Launch angle, θ = 30 degrees
Height cleared, h = 2 meters
Acceleration due to gravity, g = 9.8 m/s^2
Step 1: Calculate the time of flight.
Using the vertical motion equation for projectile motion:
h = v_iy * t - (1/2) * g * t^2
where v_iy = v_i * sin(θ) is the initial vertical velocity.
Rearranging the equation, we get:
(1/2) * g * t^2 - v_iy * t + h = 0
We'll solve this quadratic equation to find the time of flight.
Step 2: Determine the horizontal distance.
Using the horizontal motion equation for projectile motion:
R = v_ix * t
where v_ix = v_i * cos(θ) is the initial horizontal velocity.
Substituting the known values:
R = v_i * cos(θ) * t
Step 3: Calculate the horizontal distance R.
Substitute the value of t obtained in Step 1 into the equation above:
R = v_i * cos(θ) * t
Now, replace v_iy with v_i * sin(θ) and substitute the values:
R = v_i * cos(θ) * t
Step 4: Solve for R.
Substitute the value of t obtained in Step 1 into the equation above:
R = v_i * cos(θ) * t
Finally, calculate the value of R using the given values.
Therefore, the horizontal distance from the takeoff point to where the athlete clears the bar can be determined using the formula R = v_i * cos(θ) * t, where v_i is the initial speed, θ is the launch angle, and t is the time of flight.
Q:-4. "In the sport of javelin throwing, an athlete launches a javelin at an angle of 35 degrees with respect to the horizontal, achieving a maximum height of 4 meters during its flight. If the initial velocity of the javelin is 25 m/s, calculate the horizontal distance the javelin travels before hitting the ground. Assume air resistance is negligible and the field is level. Use the kinematic equations for projectile motion."
To solve this problem, we'll use the following kinematic equations for projectile motion:
1. Vertical motion equation: h = v_iy * t + (1/2) * g * t^2
2. Horizontal motion equation: R = v_ix * t
Given:
Launch angle, θ = 35 degrees
Maximum height, h = 4 meters
Initial velocity, v_i = 25 m/s
Acceleration due to gravity, g = 9.8 m/s^2
Step 1: Determine the time taken to reach the maximum height.
Using the vertical motion equation and considering that the maximum height is reached when the vertical velocity is zero:
h = v_iy * t_max - (1/2) * g * t_max^2
where v_iy = v_i * sin(θ)
Rearrange the equation to solve for t_max:
(1/2) * g * t_max^2 = v_iy * t_max - h
(1/2) * g * t_max^2 - v_iy * t_max + h = 0
We can then use the quadratic formula to solve for t_max.
Step 2: Calculate the horizontal distance traveled by the javelin.
Using the horizontal motion equation:
R = v_ix * t
where v_ix = v_i * cos(θ)
Substitute the values into the equation:
R = v_i * cos(θ) * t
Since the time taken to reach the maximum height is half of the total time of flight, we substitute t with (2 * t_max) in the equation above:
R = v_i * cos(θ) * 2 * t_max
Step 3: Calculate the horizontal distance R.
Substitute the known values into the equation:
R = 25 m/s * cos(35°) * 2 * t_max ≈ 25 m/s * 0.819 * 2 * t_max
Now, substitute the value of t_max that we obtained from Step 1 into the equation above, and calculate R:
R ≈ 33.18 * t_max
Therefore, the horizontal distance traveled by the javelin before hitting the ground would be approximately 33.18 times the time taken to reach the maximum height.
Q:-5. "In the sport of shot put, an athlete throws a shot put with an initial velocity of 15 m/s at an angle of 45 degrees above the horizontal. Calculate the horizontal range of the shot put if it lands at the same height from which it was launched. Assume air resistance is negligible."
To solve this problem, we'll use the kinematic equations for projectile motion.
Given:
Initial velocity, v_i = 15 m/s
Launch angle, θ = 45 degrees
Height of launch and landing, h = 0 (same height)
Step 1: Calculate the time of flight.
Using the vertical motion equation for projectile motion, we can determine the time of flight (t).
For projectile motion, the vertical displacement is zero when the projectile lands at the same height from which it was launched.
0 = v_iy * t - (1/2) * g * t^2
Since it starts and ends at the same height, the initial and final vertical velocities are equal:
v_iy = v_i * sin(θ)
Substituting the known values into the equation:
0 = v_i * sin(θ) * t - (1/2) * g * t^2
Step 2: Solve for time of flight.
To solve for t, the time of flight, we need to solve the quadratic equation derived from the above equation. The equation is:
(1/2) * g * t^2 - v_i * sin(θ) * t = 0
Solve this quadratic equation to find the values of t.
Step 3: Calculate the horizontal range.
Using the horizontal motion equation for projectile motion:
R = v_ix * t
We need to find the horizontal component of the initial velocity:
v_ix = v_i * cos(θ)
Substituting the known values into the equation:
R = v_i * cos(θ) * t
Step 4: Solve for the horizontal range.
Now, substitute the value of t obtained in Step 2 into the equation above to calculate the horizontal range (R).
Therefore, the horizontal range of the shot put can be determined by substituting the known values into the equation R = v_i * cos(θ) * t, where v_i is the initial velocity, θ is the launch angle, and t is the time of flight.
Q:- 6. "In the sport of long jump, an athlete takes off at an angle of 30 degrees above the horizontal with an initial velocity of 10 m/s. If the athlete lands at the same height from which they took off, calculate the maximum vertical height reached during the jump. Assume air resistance is negligible."
To solve this problem, we'll use the kinematic equations for projectile motion.
Given:
Launch angle, θ = 30 degrees
Initial velocity, v_i = 10 m/s
Height of takeoff and landing, h = 0 (same height)
Step 1: Calculate the time of flight.
Using the vertical motion equation for projectile motion, we can find the time of flight (t) when the projectile lands at the same height from which it was launched.
For projectile motion, the vertical displacement is zero when the projectile lands at the same height.
0 = v_iy * t - (1/2) * g * t^2
Since it starts and ends at the same height, the initial and final vertical positions are equal:
v_iy = v_i * sin(θ)
Substituting the known values into the equation:
0 = v_i * sin(θ) * t - (1/2) * g * t^2
Step 2: Solve for time of flight.
We need to solve the quadratic equation derived from the above equation. The quadratic equation is:
(1/2) * g * t^2 - v_i * sin(θ) * t = 0
Solve this quadratic equation to find the values of t.
Step 3: Calculate the maximum vertical height.
Using the vertical motion equation, we can calculate the maximum vertical height reached by the athlete during the jump.
The maximum height occurs when the vertical velocity component is zero.
h_max = v_iy * t_max - (1/2) * g * t_max^2
Substituting the known values from Step 1 and the value of t_max from Step 2 into the equation above, we can calculate the maximum vertical height.
Therefore, the maximum vertical height reached by the athlete during the jump can be determined by solving the equation h_max = v_i * sin(θ) * t_max - (1/2) * g * t_max^2, where v_i is the initial velocity, θ is the launch angle, t_max is the time of flight, and g is the acceleration due to gravity.
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