Question

If a freely falling body from rest covers h distance in t seconds , how much time will it require to cover 4h distance ?​

Answer 1

given:

distance covered initially = h metres

time required initially = t seconds

initial velocity(u) = 0 m/s

acceleration=gravity = 10m/s

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•applying 2nd law of motion,

$\boxed{\red{\tt s=ut +1/2at^2}}$

$\tt -h =1/2(10)t^2$

$\boxed{\tt -h =5t^2}$............(1)

•applying 1st law of motion,

$\boxed {\blue{\tt v=u+at}}$

$\tt v=10×t$

$\tt v=10t$

•let the body cover 4h distance now,

u = 0

h = 4h

t = ?

a = 10

applying 2nd law of motion,

s=ut+1/2at²

4h= 0+ 1/2(10)T²

4h=5T²

4h = 5T².....................................(2)

comparing equation (1) and (2)

5t²/5T²=1/4

t²/T²=1/4

t/T=1/2

T=2t

so the time taken to cover 4h distance is 2t time

Answer 2

Answer:

so the time taken to cover 4h distance is 2t time

Explanation:

given:

distance covered initially = h metres

time required initially = t seconds

initial velocity(u) = 0 m/s

acceleration=gravity = 10m/s

_____________________________

_____________________________

•applying 2nd law of motion,





............(1)

•applying 1st law of motion,







•let the body cover 4h distance now,

u = 0

h = 4h

t = ?

a = 10

applying 2nd law of motion,

s=ut+1/2at²

4h= 0+ 1/2(10)T²

4h=5T²

4h = 5T².....................................(2)

comparing equation (1) and (2)

5t²/5T²=1/4

t²/T²=1/4

t/T=1/2

T=2t

so the time taken to cover 4h distance is 2t time