Question

a dense ball is kicked horizontally from a height of 5 meters above the ground with an initial velocity is 20 m/s. the ball travels a horizontal distance d in time t before hitting the ground. the ball is then kicked horizontally from a bridge that is 20 m high at the original speed (20 m/s). how will the new time of flight and horizontal distance compare to the original kick? a time of flight 2t horizontal distance 2d b time of flight t horizontal distance 2d c time of flight t horizontal distance d d time of flight 2t horizontal distance d

Explanation:

Step1: Calculate time of flight

Use the formula \(h = v_{0y}t+\frac{1}{2}gt^{2}\). Initially, \(v_{0y} = 0\) (horizontal kick). For \(h = 5m\), \(h=\frac{1}{2}gt^{2}\), so \(t=\sqrt{\frac{2h}{g}}\). For \(h = 20m\), \(t'=\sqrt{\frac{2\times20}{g}} = 2\sqrt{\frac{2\times5}{g}} = 2t\).

Step2: Calculate horizontal distance

Horizontal distance \(D=v_{0x}t\) (since \(v_{0x}\) is constant, \(v_{0x} = 20m/s\)). New horizontal distance \(D'=v_{0x}t'\). Substitute \(t' = 2t\), we get \(D'=v_{0x}\times2t=2D\).

Answer:

A. Time of Flight: \(2T\), Horizontal Distance: \(2D\)