Question

A Particle is projected upwards from ground with a speed u at an angle theta with the horizontal. At the same time another paricle is dropped from a height h if they collide at a horizontal distance s from the point of projection then h is?

Answer 1

At the moment of collision
t = s/v1x , where v1x - initial horizontal component of velocity of the first object
for the second object
y(t) = h - gt^2/2 = h - gs^2/(2 u^2(cos theta)^2), but this is true only if the collision occurs before the first object rises its maximum height.

Answer 2

it then falls and its motion is described by another equation

Answer 3

Sorry I got little confused.That's all ok, we are not considering vertical motion of the first object, so the resultant equation is true for every moment of time.
At the moment of collision
t = s/v1x , where v1x - initial horizontal component of velocity of the first object
v1x = u cos theta
for the second object
y(t) = h - gt^2/2 = h - gs^2/(2 u^2(cos theta)^2)

Answer 4

@ArchiePhysics so wat is ur answer!

Answer 5

The last equation is the answer
y(t) = h - gs^2/(2 u^2(cos theta)^2)
where y - is the height at which the second object is located
h - is the initial height of it

Answer 6

Do you have any questions?