Posted by **Rajan** on Thursday, August 2, 2012 at 4:40am.

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

- Dynamics -
**MathMate**, Thursday, August 2, 2012 at 7:42am
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.

- Dynamics -
**Anonymous**, Sunday, July 21, 2013 at 1:41pm
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