Dynamics
posted by Rajan on .
A stone is thrown vertically upward from the top of a 30m high building with a velocity of 15m/s. Taking the acceleration of stone as 9.81 m/s2. And taking that as constant, determine a) the velocity v and elevations sy of stone above the ground at any time t b) the maximum altitude reached by the stone c) time when the stone strikes the ground

The kinematics equations for the stone, ignoring air resistance, is as follows:
vi=initial velocity = 15 (upwards)
g=acceleration due to gravity=(9.81)
xi=initial position = 20 m (above ground)
t=time in seconds from throwing stone upwards.
(a)
x(t)=vi*t+(1/2)gt²
(remember that g is negative)
(b) equate x'(t)=0 and solve for t.
(c) solve for x(t)=0, retain positive root only. 
Answer