Advanced AP Physics 1 Exam Question on Displacement

A ball is launched from the ground with an initial velocity of 20 m/s and at an angle of 30 degrees above the horizontal. The ball reaches a maximum height before landing on the ground again.

a) Calculate the total time of flight for the ball. b) Determine the maximum height above the ground reached by the ball. c) Find the horizontal displacement of the ball from its initial position.

Provide your answers with step-by-step detailed explanation.

Solution:

a) To calculate the total time of flight, we need to find the time it takes for the ball to reach its maximum height and then double that value.

Using the given information: Initial velocity, V₀ = 20 m/s Launch angle, θ = 30 degrees

Since the motion takes place only in the vertical direction, we can use the vertical component of velocity, Vy, to analyze the motion.

Vy = V₀ * sin(θ) Vy = 20 m/s * sin(30 degrees) Vy = 20 m/s * 0.5 Vy = 10 m/s

To find the time it takes for the ball to reach maximum height, we can use the formula: Vy = V₀y - gt where g is the acceleration due to gravity (-9.8 m/s²) and V₀y is the initial vertical velocity.

0 = 10 m/s - 9.8 m/s² * t 9.8 m/s² * t = 10 m/s t = 10 m/s / 9.8 m/s² t ≈ 1.02 s

Since the time to reach maximum height is the same as the time to fall back to the ground, the total time of flight will be twice the time to reach maximum height: Total time of flight = 2 * t Total time of flight = 2 * 1.02 s Total time of flight ≈ 2.04 s

Therefore, the total time of flight for the ball is approximately 2.04 seconds.

b) To calculate the maximum height reached by the ball, we can use the formula for vertical displacement: Δy = V₀y * t + 0.5 * g * t²

Using the values: Δy = 10 m/s * 1.02 s + 0.5 * (-9.8 m/s²) * (1.02 s)² Δy = 10.2 m - 5.0996 m Δy ≈ 5.10 m

Therefore, the maximum height above the ground reached by the ball is approximately 5.10 meters.

c) The horizontal displacement of the ball can be found using the formula: Δx = V₀x * t

To find V₀x, the initial horizontal component of velocity, we can use the equation: V₀x = V₀ * cos(θ)

V₀x = 20 m/s * cos(30 degrees) V₀x = 20 m/s * √3/2 V₀x = 10 m/s * √3

Using the total time of flight calculated earlier: Δx = 10 m/s * √3 * 2.04 s Δx ≈ 34.18 m

Therefore, the horizontal displacement of the ball from its initial position is approximately 34.18 meters.

Answer: a) The total time of flight for the ball is approximately 2.04 seconds. b) The maximum height above the ground reached by the ball is approximately 5.10 meters. c) The horizontal displacement of the ball from its initial position is approximately 34.18 meters.