Question

PLEASE HELP ASAP

David drops a soccer ball off a building. The building is 75 meters tall. (a) What is the height of the ball to the nearest tenth of a meter exactly 1 seconds after he drops the ball? (b) How many seconds, after the ball is released, will it hit the ground?

Answer 1

(a) 70.1 m

The height of the ball at time t is described by the following equation:

`h(t) = h_0 - (1)/(2)gt^2` (1)

where

`h_0 = 75 m` is the initial height of the ball

`g=9.8 m/s^2` is the acceleration due to gravity

t is the time

Since we want to know the heigth of the ball after 1 second, we just need to substitutite t=1 s into the formula, and we find:

`h(1 s)=75 m - (1)/(2)(9.8 m/s^2)(1 s)^2=70.1 m`

(b) 3.9 s

To find the time it takes for the ball to reach the ground, we have to find the time t at which the height of the ball h(t) is zero: h(t) =0. Using eq.(1), this means:

`0=h_0 - (1)/(2)gt^2`

And re-arranging we find:

`t=\sqrt{(2h_0)/(g)}=\sqrt{(2(75 m))/(9.8 m/s^2)}=3.9 s`