Question

Two stones are thrown vertically upwards with the same velocity 49m/s. If they are thrown one after the other with a time lapse of three seconds, what is the height at which they will collide ?

Answer 1

Let's say they collide after time t seconds of throwing first stone. So, we have following equation with bold letters being vectors:

r1 = ut + 0.5gt^2

r2 = u(t-3) + 0.5g(t-3)^2

Equating these two equations:

3u = 0.5g[(t-3)^2 - t^2] = -0.5g[(2t-3)(3)] => u = -0.5g(2t-3)

Putting values and assuming positive y-axis in vertically upward direction, we have,

49 j = 0.5*9.8*(2t-3) j

or, (2t-3) = 10 or t = 6.5

Hence, distance = 49*6.5 - 0.5*9.8*6.5^2 = 111.475 m (Answer)

Answer 2

Let's say they collide after time t seconds of throwing first stone. So, we have following equation with bold letters being vectors:
r1 = ut + 0.5gt^2
r2 = u(t-3) + 0.5g(t-3)^2

Equating these two equations:
3u = 0.5g[(t-3)^2 - t^2] = -0.5g[(2t-3)(3)] => u= -0.5g(2t-3)

Putting values and assuming positive y-axis in vertically upward direction, we have,

49 j = 0.5*9.8*(2t-3) j
or,  (2t-3) = 10 or t = 6.5

Hence, distance = 49*6.5 - 0.5*9.8*6.5^2 = 111.475 m (Answer