Question

from the top of a building 265 ft high a stone was dropped and at the same moment another stone was vertically thrown up with a velocity of 96fts . When and where would the two stones meet?​

Answer 1

Answer:

Let the two stones meet at a distance of x mfrom the top of the tower, and 't' be the time taken. Let us assume the downward direction as positive.

For the stone that is dropped its initial velocity u=0ms−1 ; displacement s = x and acceleration = acceleration due to gravity (g).

Using s=ut+21at2,wegetx=(0)t+21gt2→(1).

For the stone that is projected vertically upwards, its initial velocity, u=−50ms−1; displacement s =-(200-x) and acceleration a =g.

using s=ut+21at2,weget−(200−x)=−50×t+1/2gt2200=50t−1/2gt2+x→(1)(2)Fromtheequations(1)and(2)wehave200=50t−1/2gt2+1/2gt2⇒200=50∴t=4s