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# if a stone is released from a 30 m high tower and at same time a stone with 25 m/s thrown vertically upward from the base of tha

if a stone is released from a 30 m high tower and at same time a stone with 25 m/s thrown vertically upward from the base of that tower then find the position at which both stones will meet and time taken by them to meet
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@deeksha

Relative distance=30 m

Relative speed= 25 m/s

Relative acceleration=0

SO use $s=ut+\frac{1}{2}at^2$

$30=25t+0$
so $t=\frac{30}{25}=\frac{6}{5}=1.2$

Now for particle released from a 30 m high tower

a=-g m/s2

and u=0

t=1.2

so use $s=ut+\frac{1}{2}at^2$

$s=0+\frac{1}{2}(-10)(1.2)^2=-7.2m$

both stones will meet  will meet 7.2 m below from the top of tower

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