Two stones are thrown vertically upwards with the same velocity 49m/s.

1.	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» 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 = 0.59.8(2t-3) j or, (2t-3) = 10 or t = 6.5 Hence, distance = 496.5 - 0.59.8*6.5^2 = 111.475 m (ANSWER)

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»

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)

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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 ?

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