Question

a body dropped from a height h with an initial speed zero, strikes the ground with a velocity 3km/h.another body of same is dropped from the same height h with an initial speed -u'=4km/h.find the final velocity of second body with which it strikes the ground......... options are a.3km/h b.4km/h. c.5km/h. d.12km/h

Answer 1

First ball: we have, v2-u2=2as
9-0=2gh.
Therefore, h=9/2g
Second ball: we have, v2-16=2gh
v2=16+(2g×9/2g)=16+9=25
Thus v=5km/h
Ans: (c) 5km/h

Answer 2

Answer:

(c). The final velocity of the second body is 5 km/h.

Explanation:

Given that,

Initial speed u= 0

Velocity $v_{1}=3\ km/h$

Height = h

Initial speed of second body $u'= 4\ km/h$

Using equation of motion

$v^2-u^2=2as$

$9-0=2\times9.8\times h$

$h= \dfrac{9}{2\times 9.8}$

For second body

$v^2-u^2=2gh$

$v^2=16+2\times9.8\times\dfrac{9}{2\times 9.8}$

$v^2= 25$

$v= 5\ km/h$

Hence, The final velocity of the second body is 5 km/h.