Question

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 38 degree from the horizontal with a velocity of 25.0 m/s. How long was the ball in the air after the punt? How far down field does the ball land? What was the maximum height of the ball?

Answer 1

Step-by-step explanation:

This is a physics problem that involves projectile motion. To solve this problem, we can use the following equations:

- `y = y0 + v0y*t + 1/2*a*t^2`

- `x = x0 + v0x*t`

where `y` is the vertical position of the ball, `y0` is the initial vertical position of the ball, `v0y` is the initial vertical velocity of the ball, `a` is the acceleration due to gravity, `t` is the time, `x` is the horizontal position of the ball, `x0` is the initial horizontal position of the ball, and `v0x` is the initial horizontal velocity of the ball.

Using the given information, we can determine the initial vertical and horizontal velocities of the ball:

- `v0x = v0*cos(theta) = 25.0*cos(38) = 19.7 m/s`

- `v0y = v0*sin(theta) = 25.0*sin(38) = 15.2 m/s`

where `theta` is the angle of the ball's trajectory.

We can also determine the initial vertical position of the ball, which is assumed to be zero:

- `y0 = 0 m`

The acceleration due to gravity is approximately 9.81 m/s^2.

To determine how long the ball was in the air, we can use the equation for the vertical position of the ball:

- `y = y0 + v0y*t + 1/2*a*t^2`

We want to find the time when the ball hits the ground, so we can set `y` equal to zero:

- `0 = 0 + 15.2*t + 1/2*(-9.81)*t^2`

Solving for `t`, we get:

- `t = 3.2 s`

Therefore, the ball was in the air for **3.2 seconds**.

To determine how far down field the ball lands, we can use the equation for the horizontal position of the ball:

- `x = x0 + v0x*t`

We know that `x0` is zero, so we can simplify the equation to:

- `x = v0x*t`

Substituting in the values we know, we get:

- `x = 19.7 m/s * 3.2 s = 63.0 m`

Therefore, the ball lands **63.0 meters** down field.

To determine the maximum height of the ball, we can use the equation for the vertical position of the ball:

- `y = y0 + v0y*t + 1/2*a*t^2`

We want to find the maximum height, so we can find the time when the ball reaches its maximum height. This occurs when the vertical velocity of the ball is zero:

- `v = v0y + a*t`

Setting `v` equal to zero and solving for `t`, we get:

- `t = v0y/a`

Substituting in the values we know, we get:

- `t = 15.2 m/s / 9.81 m/s^2 = 1.55 s`

We can now use this time to find the maximum height:

- `y = y0 + v0y*t + 1/2*a*t^2`

Substituting in the values we know, we get:

- `y = 0 + 15.2 m/s * 1.55 s + 1/2*(-9.81 m/s^2)*(1.55 s)^2 = 11.3 m`

Therefore, the maximum height of the ball is **11.3 meters**.