Question

A marble is dropped from the top of a tower 150m high and at the same time another ball is projected vertically upwards from the ground with a velocity of 25m/sec. Calculate where and when the two stones will meet. Take g=10m/s2.

Answer 1

Explanation:

take the height of meeting point as x

ball thrown down:

s=ut+1/2at^2

(100-x)= 0+1/2 ×10×t^2

(100-x)= 5×t^2 ---------------------------–1

ball thrown down:

s=ut+1/2at^2

x= 25t+1/2(-10)t^2

x=25t+(-5)t^2 ---------------------------–2

1&2

100-x+x= 25t +5t^2-5t^2

100= 25t

t= 4

take any one of 1 or 2

and substute