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A ball is thrown vertically upward with a velocity of 22m/s from a point 4.6m above the ground.Find the velocity with which it hits the ground?If it rebounds with a velocity which is half the velocity it hits the ground,Find the greatest height after the bounce?

Vidura , 8 Years ago
Grade 11
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

From point 4.6 m, with velocity 22 m/s,
maximum height it travels = u^2/2g = 22*22/20 = 24.2 m
that means
at total height if we includes height of building too,
from 4.6m + 24.2 m =  28.8m we will apply 2nd law of eqn.
therfore,
from 28.8 m velocity of ball =0
hence final velocity at ground,
v^2 = 0^2 + 2*g*28.8
v^2 = 576
v = 24 m/s
 
  1.  If it rebounds with a velocity which is half the velocity it hits the ground,
.i.e  Greatest height, H = (v/2)^2/2g  = 144/20 = 7.2 m

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