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

Asked by Shourya Mukherjee | 10th May, 2012, 10:03: PM

Expert Answer:

 
apply, to find initial velocity.
v2-u2=2gs
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

Answered by  | 11th May, 2012, 02:40: PM

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