Question

A stone is dropped from the top of the building and at the same time a second stone is is thrown vertically upward with a speed of 20m/s. if height of the building is 60m. Find the time after which the stones collide. stones are moving in the same line

Answer 1

Let the distance they meet at be x

Total journey = 60m
Distance at which first stone meet = x (since it will travel only x metres)

For the first stone,
u = 0m/s ( since it's dropped freely)
t = unknown so let it be t
a = 10m/s² ( acceleration due to gravity)


By using s = ut + 1/2at² we get,

x = 0 × t + 1/2 × 10 × t²

=> x = 5t².............. equation i


For second stone,
u = 20m/s (given)
t = unknown so let it be t (since both are projected at the same time)
g = -10m/s² (acceleration due to gravity acting in opposite direction)


Since the first stone will collide at x m, the displacement of second stone will be 60 - x


So we get,
for second stone
60 - x = ut + 1/2at²

=> 60 - x = 20t + 1/2(-10)t²

=> 60 - x = 20t - 5t². equation ii



Adding equation i and ii we get

60 - x + x = 20t - 5t² + 5t²

=> 60 = 20t

=> t = 60/20

=> t = 3 second


3 seconds is your answer.


Also if you want to know the distance at which they will meet then

x = 5t²

=> x = 5 × 3²

=> x = 5 × 9

=> x = 45 m

They will collide at 45 m from the top at 3 seconds

Hope it helps dear friend ☺️✌️✌️