Question #d18ed

1 Answer
Jun 27, 2015

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 hh, the only parameter that influences that object's time of flight is the actual height, i.e. the value of hh.

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