Question

In a baseball game a foul ball was hit straight up with a speed if 30 m/s. Ignore air resistance.

a. How high did the ball go up?
b. How long was the ball in the air?
c. What happened to the answers in part “a” and “b” if the initial velocity was only 15 m/s?

initial position = ?
finial position = ?
initial velocity = ?
final velocity = ?
acceleration = ?
time = ?

Answer 1

a. At maximum height h, the baseball has zero vertical velocity, so that with initial velocity v we have

`0^2-v^2 = -2gh \implies h = (v^2)/(2g)`

Given that
`v = 30 (\rm m)/(\rm s)`, the ball reaches a height of

`h = (\left(30(\rm m)/(\rm s)\right)^2)/(2g) \approx \boxed{46\,\mathrm m}`

b. At max height, the baseball's vertical velocity is such that

`0=v-gt`

For
`v = 30 (\rm m)/(\rm s)`, we find

`gt=30(\rm m)/(\rm s) \implies t = \frac{30(\rm m)/(\rm s)}g \implies t \approx 3.1 s`

which is half the time it spends in the air, making the total time about 6.1 s.

c. We saw in part (a) that

`h = (v^2)/(2g)`

If we halve the initial speed and replace
`v` with
`\frac v2`, we get

`h = (\left(\frac v2\right)^2)/(2g) = \frac14 \cdot (v^2)/(2g)`

which means the max height is reduced by a factor of 1/4.

In part (b), we found the time to max height is

`t = (v)/(g)`

and halving the initial speed gives

`t = \frac{\frac v2}g = \frac12 \cdot (v)/(g)`

That is, the time to max height is reduced by a factor of 2 when the speed is halved, and so the overall time in the air is reduced by a factor of 2.

Answer 2

h=4+85*time - 1/2 32*time^2

16time^2-85*time-106=0

use the quadratic formula to solve for time (notice two solutions, one going up, one going back down).

time=(85+-sqrt(85^2+4*16*106) )/32

Step-by-step explanation:

Recheck my work just in case I'm not good at this type of class but I wanted to help.