Thu May 10, 2012 By: Shourya Mukherjee

A body projected vertically upwards crosses point A and B separated by 28m, with velocities one-third and one-fourth of the initial velocity respectively. What is the maximum height reached by it above the ground? Shourya

Expert Reply
Fri May 11, 2012
apply, to find initial velocity.
here we have taken v as initial velocity
v2/16=v2/9  -2x10x 28
v=48?5 m/s
now to find max ht:
v=o at max height
so 0=v-2 x 10 x h
h=576m by putting v=48?5 m/s
Distance = 576 m
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