Question

A stone is allowed to fall down 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.​

Answer 1

Answer:

4s and 21.6m from the ground

For the falling stone ,

H=100m

a=0m/s

For the stone thrown downwards,

u=25m/s

a=-gm/s^2

Let the two stone meet at time t s after the start and the distance x m from the ground.

(100-x)=ut+1/1gt^2

0+1/2gt^2-----------(1)

F

FOR THE STONE THROWN UPWARDS

x=25×t-1/2gt^2=25×t-1/2gt^2

100=25t

t=4s

THUS ,RHE TWO STONES MEET 4SECOND FROM THE START

FROM EQ1

• X=25M/S×4S-1/2gt^2×9.8m/s^2×4^2s^2
• =100m-78.4m=21.6m

U can use ut+1/2gt^2 instead of UT-1/2gt^2

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