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 #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#