If two bodies are projected from the top of a tower in opposite directions with velocities of u₁ and u₂ simultaneously, the time when their velocities are mutually perpendicular is t = √(u₁u₂)/g
If two bodies are projected from the top of a tower in opposite directions with velocities of u₁ and u₂ simultaneously. To find the time when their velocities are mutually perpendicular, we proceed as follows
Since the bodies have initial velocities u₁ and u₂, their velcities at any point in vector form is using the equation of motion under gravity, v = u - gt
For body 1 v₁ = u₁ - gt
For body 2 v₂ = u₂ - gt
Now for the velocities of the bodies to be mutually perpendicular, their dot product must be zero.
So, v₁.v₂ = 0
(u₁ - gt).(u₂ - gt) = 0
u₁.u₂ - u₁.gt + u₂.(- gt) + (- gt).(-gt) = 0
u₁.u₂ - u₁.gt - u₂.(gt) + (gt).(gt) = 0
Now,
- u₁.u₂ = u₁u₂cos180 = -u₁u₂(since they are projected in opposite directions and parallel and the angle between them is zero)
- u₁.gt = 0 (since they are perpendicular)
- u₂.(gt) = 0 (since they are perpendicular) and
- (gt).(gt) = (gt)(gt)cos0 = g²t² (since they are parallel and the angle between them is zero)
Thus
u₁.u₂ - u₁.gt - u₂.(gt) + (gt).(gt) = 0
-u₁u₂ - 0 - 0 + g²t² = 0
-u₁u₂ - 0 + g²t² = 0
-u₁u₂ + g²t² = 0
-u₁u₂ = -g²t²
Dividing both sides by -g², we have that
-u₁u₂/-g² = -g²t²/-g²
t² = u₁u₂/g²
Taking square root of both sides, we have that
t² = u₁u₂/g²
√t² = √(u₁u₂/g²)
t = √(u₁u₂)/g