Question

small platform starts moving from rest from ground in upward direction with uniform acceleration 5 m/s^2 at time t is equal to 2 seconds a stone is projected horizontally in frame of platform with speed 10 metre per second. the maximum height obtained by the stone with respect to the ground is

Answer 1

HELLO DEAR,

The height attained by the platform = (1/2)*at^2

=(1/2)*5*2^2

=10

Now the height gained by projectile from the frame of reference of platform=u^2/2*g

              =10^2/2*10

              =5m

Hence total height attained=10+5

             =15m

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

platform starts moving from rest from ground in upward direction with uniform acceleration 5m/s². after 2 second vertical velocity of platform is v.
use formula, v = u + at
v = 0 + 5 × 2 = 10 m/s

vertical velocity of platform = vertical velocity of stone = 10 m/s

horizontal velocity of stone = 10 m/s

we know from projectile motion,
maximum height with respect to platform = {vertical component }²/2g
= 10²/(2 × 10)
= 5m

hence, maximum height with respect to ground = maximum height with respect to platform + height of platform with respect to ground
= 5 + 10 = 15m