Question

David drops a ball from a bridge at an initial height of 100 meters.

(a) What is the height of the ball to the nearest tenth of a meter exactly 3 seconds after he releases the ball?

(b) How many seconds after the ball is released will it hit the ground?

Answer 1

(a) 55.9 m

(b) About 4.518 seconds.

_____
A graphing calculator can be helpful for these.

Answer 2

The general formula is:

`y = y_(o) + v_(o)t - (1*g*t^(2) )/(2)` -- (A)

Where g = 9.8 m/s/s

`v_(o)` = 0

`y_(o)` = 100

a)
Time=t=3s.
Plug-in the values in (A)
y = 100 -(1/2*9.8*3*3)
y = 100-44.1 = 55.9m

Height of the ball exactly after 3 seconds = 55.9m.

b)
At ground, y=0; Plug-in values in (A):
A=> 0 = 100 - (1/2* 9.8 * [tex]t_{2}[/text])
Therefore t = + 4.52 s and - 4.52 seconds.

As t cannot be negative, therefore:
Time when ball hits the ground = 4.52s

-i