Question

A stone is allowed to fall from the top of a tower 100m
and at the same time another stone is projected vertically
upwards from the ground with a velocity of 25 m/s. Calculate
when and where the two stones will meet.​

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

Let s1 be distance fallen by freely falling stone when two stones met at time t

s1= ut+ 1/2gt^2 u=0

s1= 1/2gt^2 .....eq1

Let s2 be distance moved by throwed stone at t

s2= ut-1/2gt^2

s2=25t-1/2gt^2 ....eq2

Add eq1 and eq2

S1+ S2 = 25t

100= 25t

t = 4 secs

s1 = 1/2g4^2

s1= 80m

Then s2 = 20m

Explanation: