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 highest point this ball will reach?

Options

A. 38.66 m
B. 11.48 m
C. 9.80 m
D. 1.53 m.

Answer

B

Explanation

At highest point V=0, `V_0`=-15m/sec, a=`9.8m//sec^2`, and t=?
`V=V_0 + at`
O=-15+9.8 t
15=9.8 t i.e. t=153
X=(-15)(1.53)+1/2 (9.8)`(1.53)^2`
X=-22.95+(4.9)(2.34)
X=-22.95+11.47
X=11.48 m above the ground