A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the two stones will meet?
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✓suppose AB is a tower with hight H=100m
stone A is dropped (u=0) and stone B is thrown upward with initial velocity (u=25m/s).
let the two stone mit at C at time t.
atA initial velocity, u=0m/s
Acceleration due to gravity,g=10m/s
so distance travelled,
[ s=ut+1/2×gt^2]
AC=0+1/2gt^2 - AC=1/2 gt^2......(1)
At B (u=25m/s)
acceleration= -10 m/s
distance=BC=(25m/s)t+1/2(-g)t^2.
BC=(25m/s)t-1/2gt^2..............(2)
by (1) and (2),
AC+BC=25m/s
AC+BC=H= m
100m=25m/s×t
t =100m/25 m/s
=4s ans...
AC=1/2×10 m/s×4s= 5m/s×16 s^2
=80 m. ans...
stone A is dropped (u=0) and stone B is thrown upward with initial velocity (u=25m/s).
let the two stone mit at C at time t.
atA initial velocity, u=0m/s
Acceleration due to gravity,g=10m/s
so distance travelled,
[ s=ut+1/2×gt^2]
AC=0+1/2gt^2 - AC=1/2 gt^2......(1)
At B (u=25m/s)
acceleration= -10 m/s
distance=BC=(25m/s)t+1/2(-g)t^2.
BC=(25m/s)t-1/2gt^2..............(2)
by (1) and (2),
AC+BC=25m/s
AC+BC=H= m
100m=25m/s×t
t =100m/25 m/s
=4s ans...
AC=1/2×10 m/s×4s= 5m/s×16 s^2
=80 m. ans...
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