A person initially at rest throws a ball upward at an angle θ0= 75 ∘ with an initial speed v0=15 m/s . He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval of 1.03 s and then continues to run at a constant speed for the rest of the trip. He catches the ball at exactly the same height he threw it. Let g= 9.81 m/s2 be the gravitational constant. What was the person's acceleration a (in m/s2)?

Question

A person initially at rest throws a ball upward at an angle θ0= 75 ∘ with an initial speed v0=15 m/s . He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval of 1.03 s and then continues to run at a constant speed for the rest of the trip. He catches the ball at exactly the same height he threw it. Let g= 9.81 m/s2 be the gravitational constant. What was the person's acceleration a (in m/s2)?

a=

Answer 1

To find the person's acceleration, we need to use the equations of motion. First, let's break down the motion of the ball into two parts: the upward motion and the downward motion.

1. Upward Motion:
During the upward motion, the only force acting on the ball is the gravitational force, which is directed downward and causes the ball to slow down until it reaches its peak height and starts to fall back down. The acceleration during the upward motion is equal to the acceleration due to gravity, g = 9.81 m/s^2.

2. Downward Motion:
During the downward motion, the ball is accelerating due to both gravity and the person's acceleration, since the person is trying to catch up to the ball. Let's assume the person accelerates with a constant acceleration, a, for a time interval of t = 1.03 s.

Now, to find the person's acceleration, we can use the following equation for displacement:

s = v0t + (1/2)at^2

Since the ball is caught at the same height it was thrown, the displacement, s, during the upward and downward motions should cancel out. Therefore, we can write two equations for the upward and downward motions:

1. Upward Motion:
0 = v0*sinθ0*t + (1/2)(-g)t^2 ... (equation 1)

2. Downward Motion:
0 = (v0*cosθ0 + at)*t + (1/2)(-g)t^2 ... (equation 2)

In equation 1, we have the initial velocity, v0 = 15 m/s, the angle of projection, θ0 = 75°, and the time interval, t = 1.03 s.

In equation 2, we have the same initial velocity, v0 = 15 m/s, the angle of projection, θ0 = 75°, the time interval, t = 1.03 s, and the person's acceleration, a, which we want to find.

Now, let's solve these two equations simultaneously to find the person's acceleration, a.