Question

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 25ms-1.calculate when and where the stones will meet.​

Answer 1

Answer:

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Explanation:

80m

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Answer 2

Explanation:

let me explain to you how

let a stone A is allowed to fall from a height H= 100m

its initial velocity u1 =0

and a1 =+g = 10ms-2

another stone B is projected upwards from the ground with an initial velocity u2= 25ms-1and for it a2= -g = -10ms-2

let the two stones meet at a point C at a distance y below a or(100-y)above B after a time t.

y= u1 t+1/2a1  t2

=0x t+1/2*10*t2

y = t t2

for stone b,(100-y) = u2 t + 1/2a2 t2

=25-5t2

adding (1) and (2)

100=25t

t=100/25=4s

substituting value of t in (1)we get

y = 5*4^2=80m

over and happy to help:)