Question

3. David drops a ball from a bridge at an initial height of 70 meters.
(a) What is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball?
(b) How many seconds after the ball is released will it hit the ground?

Answer 1

for the first one you need the equation
s=ut+at^2/2


s is distance travelled u= initial velocity (=0) t = time = 2
so this gives you distance travelled - since it started at 70m height you can then work out its height after 2 s

the scond one uses the SAME equation, but this time
s=70 u=0 so find t

in both cases acceleration a = gravity = g