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)?

To find the person's acceleration, we can first find the time it takes for the ball to reach its maximum height and then find the time it takes for the person to catch up to the ball at that height.

1. Let's find the time it takes for the ball to reach its maximum height. We can use the equation for vertical motion:

vf = vi + gt

where vf is the final velocity, vi is the initial velocity, g is the acceleration due to gravity, and t is the time.

At maximum height, the final velocity will be zero. So, we have:

0 = 15sin(75°) - 9.81t

Solving for t, we find:

t = (15sin(75°)) / 9.81

2. Now, let's find the time it takes for the person to catch up to the ball at the maximum height. The person is accelerating for a time interval of 1.03 seconds and then runs at a constant speed for the rest of the trip. Let's call the time it takes for the person to reach the maximum height t1.

From the equation of motion for constant acceleration:

d = vit + (1/2)at^2

Since the person catches the ball at the same height, the initial and final displacements are both zero.

0 = 0 - (1/2)at1^2

Solving for t1, we find:

t1 = sqrt((2*0) / a) = 0

This means that the person reaches the maximum height in zero seconds when using a constant acceleration.

3. To find the total time taken by the person to reach the maximum height, we add the time it takes for the ball to reach the maximum height and the time it takes for the person to catch up to the ball at that height:

total time = t + t1 = (15sin(75°)) / 9.81

4. Now, we know that the person runs at a constant speed for the rest of the trip. Let's call this speed v.

The horizontal distance traveled by the person during the time taken to reach the maximum height is equal to the horizontal distance traveled by the ball during the same time. This can be calculated using:

distance = velocity * time

v * total time = 15cos(75°) * total time

5. Since the person catches the ball at the same height, the time taken for the person to reach the maximum height is equal to the time taken for the person to descend from the maximum height to the ground.

Using the equation of motion for vertical motion:

d = vit + (1/2)at^2

We know the initial velocity is zero, and the distance is equal to the height the ball was thrown.

h = 0 + (1/2)(-9.81)t^2

Rearranging the equation, we find:

t^2 = (2h) / 9.81

Substituting the value of h, we find:

t^2 = (2 * h) / 9.81

Solving for t, we find:

t = sqrt((2 * h) / 9.81)

6. The total time taken by the person to reach the maximum height and descend back to the ground is:

total time = t + t = 2t = 2 * sqrt((2 * h) / 9.81)

7. Now, we can find the constant acceleration a using the equation:

distance = (1/2)at^2

Rearranging the equation, we have:

a = (2 * distance) / t^2

Substituting the value of distance, which is equal to the horizontal distance traveled by the person:

a = (2 * (v * total time)) / t^2

Substituting the value of total time and t:

a = (2 * (v * (2 * sqrt((2 * h) / 9.81)))) / ((2 * h) / 9.81)

Simplifying the equation, we get:

a = (4 * v * sqrt((2 * h) / 9.81)) / h

Therefore, the person's acceleration a is (4 * v * sqrt((2 * h) / 9.81)) / h.

To find the person's acceleration, we need to consider the time taken to reach the same height as the ball when thrown upward.

Let's break down the problem into different parts:

1. Vertical motion of the ball:
When throwing the ball upward, it experiences a vertical motion influenced by gravity. We can use the equations of motion to calculate the time it takes for the ball to reach the same height at which it was thrown.

The ball's initial vertical velocity, V_y0, is given by:
V_y0 = v0 * sin(θ0)

The time taken for the ball to reach its maximum height can be found using the equation:
t_max = V_y0 / g

2. Catching up with the ball:
The person accelerates with a constant acceleration, a, for a time interval of 1.03 seconds. During this time, the person is decreasing the vertical distance between themselves and the ball.

Using the equations of motion, we can find the change in height, Δy, for the person during this time:
Δy = (1/2) * a * t^2

Where t is the time interval (1.03 seconds).

3. Running at a constant speed:
After accelerating, the person maintains a constant speed until they catch the ball. During this time, the person's vertical displacement is equal to the distance they decreased while accelerating.

Using the equation for constant speed motion, we can find the person's displacement over this time:
s = v * t

Where s is the displacement, v is the constant velocity, and t is the time.

4. Equating the displacement of the person and the ball:
The person's displacement during the constant speed phase should be equal to the displacement they decreased during acceleration. That is:
Δy = s

Now we have all the necessary information to solve for the person's acceleration.

First, find the time taken for the ball to reach its maximum height, using the given values of θ0 and v0.

Next, calculate the displacement of the person during the constant speed phase using the value of t obtained in the previous step.

Finally, equate the displacement of the person with the decrease in height during acceleration and solve for the acceleration, a.

I hope this explanation helps you solve the problem!

Vo = 15m/s[75o]

Xo = 15*cos75 = 3.88 m/s.

Range = Vo^2*sin(2A)/g
Range = 15^2*sin(150)/9.81 = 11.47 m.

d = 0.5a*t^2 = 11.47 m.
0.5a*1.03^2 = 11.47
0.53a = 11.47
a = 21.6 m/s^2.