Example 1. 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 m/s?, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.
Given: m = 3 kg, distance travelled by the
ball s = 125 m, initial velocity of the ball =
u=o and acceleration a = g = 10 m/s2.
(i) Newton's second equation of motion
gives
Answers
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:
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=93.75
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
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