Question

an iron ball of mass 3 kg is released from a height of 125 M and falls freely to the ground. Assuming that the value of g is 10 metre per second square, calculate a) time taken by the ball to reach ground? b)the velocity of the ball on reaching the ground? c) the height of the ball at half time it takes to reach the ground

Answer 1

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




I hope this will help you

Answer 2

Answer:5sec

Explanation:

By the formula of

s=ut+1/2gt(square)

and take the value of g=-10