Question

Question #d18ed

Answer 1

The ratio of the time of flight for the two balls will be equal to 1.

Explanation:

This is a great example of a trick question.

When you launch an object horizontally from a height `h`, the only parameter that influences that object's time of flight is the actual height, i.e. the value of `h`.

The launch velocities have no impact on the total time of flight, they only affect the horizontal distance covered by the object.

The movement of the two balls has a horizontal component and a vertical component. The same can be said for the initial velocity.

`{(v_(0x) = v_0 * cos(theta)), (v_(0y) = v_0 * sin(theta)) :}`

When you launch a ball horizontally, the angle it makes with the horizontal is equal to `0^@`, which implies that

`{ (v_(0x) = v_0 * cos(0^@) = v_0) ,(v_(0y) = v_0 * sin(0^@) = 0) :}`

Since the initial velocity has no vertical component, the vertical displacement will be equal to

`-h = underbrace(v_(0y))_(color(blue)("=0")) * t - 1/2 g t^2`

`h = 1/2 g t^2`, where

`h` - the height of the tower;
`t` - the total time of flight.

This means that you have

`h = 1/2 g * t_1^2` `->` for the first ball;

`h = 1/2 g * t_2^2` `->` for the second ball;

Therefore,

`cancel(1/2 * g) * t_1^2 = cancel(1/2 * g) * t_2^2`

`t_1^2 = t_2^2 => t_1/t_2 = sqrt(1) = 1`