Question

Two bodies are projected horizontally in
opposite directions with velocities u1 and u2
from the top of a building. If their velocities
become perpendicular to each other after
falling through distance h, then h=?​

Answer 1

Answer:

h = u1u2/2g

Explanation:

after falling through distance h, h = gt²/2

∝ = 0

for particle 1

vx = u1cos∝ = u1

vy = u1sin∝+gt = gt

for particle 2

vx = u2cos∝ = u2

vy = u2sin∝+gt = gt

tanФ1 = vy/vx = gt/u1

tanФ2 = vy/vx = gt/u2

their velocities  become perpendicular to each other after

tanФ1tanФ2 = 1

g²t²/u1u2 =1

2g²t²/2u1u2 = 1

2gh = u1u2

h = u1u2/2g