Question

A baseball is popped straight up into the air and has a hang-time of 6.25 S.

Determine the height to which the ball rises before it reaches its peak. (Hint: the
time to rise to the peak is one-half the total hang-time.)

Answer 1

To determine the height to which the ball rises before it reaches its peak, we need to know the initial velocity of the ball and the acceleration due to gravity. Let's assume the initial velocity of the ball is v and the acceleration due to gravity is g.

The time it takes for the ball to reach its peak is one-half the total hang-time, or 1/2 * 6.25 s = 3.125 s.

The height to which the ball rises can be calculated using the formula:

height = v * t - (1/2) * g * t^2

Substituting in the values we know, we get:

height = v * 3.125 s - (1/2) * g * (3.125 s)^2

To solve for the height, we need to know the value of v and g. Without more information, it is not possible to determine the height to which the ball rises before it reaches its peak.

Step-by-step explanation:

Answer 2

Approximately
`47.9\; {\rm m}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}` and that air resistance on the baseball is negligible.)

Step-by-step explanation:

If the air resistance on the baseball is negligible, the baseball will reach maximum height at exactly
`(1/2)` the time it is in the air. In this example, that will be
`t = (6.25\; {\rm s}) / (2) = 3.125\; {\rm s}`.

When the baseball is at maximum height, the velocity of the baseball will be
`0`. Let
`v_(f)` denote the velocity of the baseball after a period of
`t`. After
`t = 3.125\; {\rm s}`, the baseball would reach maximum height with a velocity of
`v_(f) = 0\; {\rm m\cdot s^(-1)}`.

Since air resistance is negligible, the acceleration on the baseball will be constantly
`a = (-g) = (-9.81\; {\rm m\cdot s^(-2)})`.

Let
`v_(i)` denote the initial velocity of this baseball. The SUVAT equation
`v_(f) = v_(i) + a\, t` relates these quantities. Rearrange this equation and solve for initial velocity
`v_(i)`:

`\begin{aligned}v_(i) &= v_(f) - a\, t \\ &= (0\; {\rm m\cdot s^(-1)}) - (-9.81\; {\rm m\cdot s^(-2)})\, (3.125\; {\rm s}) \\ &\approx 30.656\; {\rm m\cdot s^(-1)}\end{aligned}`.

The displacement of an object is the change in the position. Let
`x` denote the displacement of the baseball when its velocity changed from
`v_(i) = 0\; {\rm m\cdot s^(-1)}` (at starting point) to
`v_(t) \approx 30.656\; {\rm m\cdot s^(-1)}` (at max height) in
`t = 3.125\; {\rm s}`. Apply the equation
`x = (1/2)\, (v_(i) + v_(t)) \, t` to find the displacement of this baseball:

`\begin{aligned}x &= (1)/(2)\, (v_(i) + v_(t))\, t \\ &\approx (1)/(2)\, (0\; {\rm m\cdot s^(-1)} + 30.565\; {\rm m\cdot s^(-1)})\, (3.125\; {\rm s}) \\ &\approx 47.9\; {\rm m}\end{aligned}`.

In other words, the position of the baseball changed by approximately
`47.9\; {\rm m}` from the starting point to the position where the baseball reached maximum height. Hence, the maximum height of this baseball would be approximately
`47.9\; {\rm m}\!`.