A stone is launched straight up by a slingshot. Its initial speed is 19.6 m/s, and the stone is 1.50 m above the ground when launched, (a) How high above the ground does the stone rise? (b) How much time elapses before the stone hits the ground?

Answer: a) 19.21m b) 3.92secs

Explanation:

a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.

Maximum height = U²/2g

U is the initial velocity = 19.6m/s

g is acceleration due to gravity = 10m/s²

Maximum Height = 19.6²/2 (10)

H = 19.21m

b) The time elapsed before the stone hits the ground is the time of flight T = 2U/g

T = 2 (19.6) / 10

T = 39.2/10

Time elapsed is 3.92secs