a baseball is hit at a height h = 1.10 m and then caught at the same height. It

travels alongside a wall, moving up past the top of the wall 1.2 s after it is hit and then down
past the top of the wall 3.2 s later, at distance D = 53 m farther along the wall. (a) What
horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and
(c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is
the wall?

To solve this problem, we need to use the equations of motion for projectile motion. Let's break it down step by step:

(a) What horizontal distance is traveled by the ball from hit to catch?

The horizontal distance traveled by the ball is given by the formula:
Distance = velocity × time

We can calculate the horizontal velocity of the ball using the information provided. Since the ball is caught at the same height as it was hit, the vertical component of the velocity is zero. We only need to consider the horizontal component.

First, let's calculate the time it takes for the ball to reach the top of the wall and then come down past the top of the wall. The time for the ball to reach the top of the wall is given as 1.2 seconds, and the time for the ball to come down past the top of the wall is given as 3.2 seconds. The total time of flight is the sum of these two times: 1.2 + 3.2 = 4.4 seconds.

Now, let's find the horizontal distance traveled. Since the vertical component of the velocity is zero, the time of flight is the same as the time it takes for the ball to travel horizontally.

Distance = horizontal velocity × time of flight

We need to find the horizontal velocity. We can use the formula:

Avg velocity = total displacement / total time

The total displacement is given as 53 meters, and the total time is 4.4 seconds.

Avg velocity = 53 m / 4.4 s

Now we have the average horizontal velocity. Multiply it by the total time to get the horizontal distance traveled:

Horizontal distance = Avg velocity × total time

(b) What is the magnitude of the ball's velocity just after being hit?

The magnitude of the velocity is given by the formula:

Velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2)

Since the vertical component of the velocity is zero, the magnitude of the velocity just after being hit is equal to the horizontal velocity.

(c) What is the angle (relative to the horizontal) of the ball's velocity just after being hit?

The angle can be found using the Tan function:

Angle = arctan(vertical velocity / horizontal velocity)

Since the vertical component of the velocity is zero, the angle will be zero degrees.

(d) How high is the wall?

To find the height of the wall, we need to use the equation for vertical motion:

h = initial vertical velocity × time + (1/2) × acceleration × time^2

Since we know the ball is caught at the same height it was hit, the vertical component of the velocity is zero at the highest point. The time of flight is 4.4 seconds, as calculated before.

We can rearrange the equation to solve for the initial vertical velocity:

initial vertical velocity = (h - (1/2) × acceleration × time^2) / time

Substitute the values given for h, acceleration, and time to get the initial vertical velocity.

#YOLO