In uniformly accelerated motira the following questions hold :
V=Vo + at
`X=Vot + 1//2 at^2`
When X = displacement , V=Velocity at time, Vo= initial velocity, t=time , and a =acceleration. A ball is projected directly upward at a velocity of 15 m/sec.
What is the distance above the ground after 3 seconds ?

Options

A. 1.8 m
B. 0.9 m
C. 0 m
D. 3.6 m.

Answer

B

Explanation

Choose the upward direction as negative and downward as positive
Then `V_0`=-15m/sec and `a=9.8 m//sec^2` at t=3 seconds, X=?
X=`V_"ot"+1//2at^2`
X=(1-15)(3)+(1/2)(9.8)`(3)^2`
=-45+44.1
X=-0.9 m above ground .