plz write out solution step by step
Q1.
Using 2nd kinematic equation:
S = U*t + 0.5*a*t^2
Using given value:
400 = U*5 + 0.5*a*5^2
divide above equation by 5
80 = U + 2.5*a
Using 1st kinematic equation
V = U + a*t
0 = U + a*5
a = -U/5
Using this value in above equation
80 = U + 2.5*a
80 = U + 2.5*(-U/5)
80 = U/2
U = 160 m/sec = initial velocity.
2.
Avg Velocity is given by:
Vavg = displacement/time taken
Given that Vavg = 6 m/sec
So displacement = (6 m/sec)*4 sec = 24 m
Now using 2nd kinematic equation:
S = U*t + 0.5*a*t^2
24 = 2*4 + 0.5*a*4^2
8*a = 16
a = 2 m/sec^2 = acceleration
3.
dropped means initial velocity = 0 m/sec
Using 2nd kinematic equation
S = U*t + 0.5*a*t^2
S = height of building = 100 m
U = 0
a = g = 9.8 m/sec^2
So,
100 = 0*t + 0.5*9.8*t^2
t = sqrt (200/9.8)
t = 4.52 sec = time taken to hit the ground
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plz write out solution step by step Irlte out your legible answers in the space provided....
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