Afoul ball is hit straight up into the air with a speed of a bout 25 m/s (a) how high does it go? (b) how long is it in the air

The relevant equations we can use in this problem are:

d = v0 t + 0.5 a t^2

v^2 = v0^2 + 2 a d

a. how high does it go?

We are to find for the distance d. The formula is:

v^2 = v0^2 + 2 a d

where v is final velocity = 0 at the peak, v0 = 25 m/s, a is negative gravity since opposite to direction = - 9.8 m/s^2, d is height achieved = ?

0 = (25)^2 + 2 (-9.8) d

d = 31.89 m

Therefore the maximum height is about 31.89 meters

b. how long is it in the air

We use the formula:

d = v0 t + 0.5 a t^2

in going up, a = -g = -9.8 m/s^2, find for time t:

31.89 = 25 t + 0.5 (- 9.8) t^2

-4.9 t^2 + 25 t = 31.89

t^2 – 5.1 t = 6.51

Completing the square:

(t – 2.55)^2 = 6.51 + 6.5025

t – 2.55 = ± 3.61

t = -1.06, 6.16

Since time cannot be negative, therefore:

t = 6.16 seconds

However this is not the total time yet, because this only accounts for the time going up.

The time going up and going down are equal therefore:

t(total) = 2 t

t(total) = 12.32 seconds

Therefore the foul ball spent 12.32 seconds in the air.

trig 15.3 cos

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