Question

an iron ball mass 3kg is released from a height of 125m and falls freely to the ground assuming that the value g=10m/S2
calculate:
(1) time nahi taken by ball to reach the ground
(2) velocity of ball on reaching the ground
(3) height of ball at half the times takes to reach the ground​

Answer 1

Answer:

S=125=UT+1/2AT^2

as U is =0

therefore S=1/2AT^2

here A is 10m/s^2

therefore

125=1/2AT^2

250=10T^2

therefore T =√25=5sec

V^2-U^2=2AS

here A is -10 and S is -125m as its falling from a height and U =0 therefore

V^2=2(-10)(-125)

V=√2500

V=50m/s

half time by it reaches =5/2=2.5sec

S=UT+1/2AT^2

S=1/2x10xT^2

therefore S=31.25m

but H=125-31.25

therefore H=93.75m

a)The time by which the ball reaches the ground is=5seconds

b) The velocity of the ball on reaching the ground=50m/s

c) The height of the ball at half time it takes to reach the ground=93.75m