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.
if the initial velocity of the ball is doubled to 40 m/s, how will the new time of flight and horizontal distance for this launch compare to the original kick?
a
time of flight: 2t
horizontal distance: 2d
b
time of flight: t
horizontal distance: 2d

Explanation:

Step1: Analyze the time - of - flight formula

The vertical motion of the ball is a free - fall motion. The formula for the vertical displacement \(y = v_{0y}t+\frac{1}{2}at^{2}\). Since the ball is kicked horizontally, \(v_{0y} = 0\), and \(y=-h\) (taking downwards as negative), \(a=-g\). So, \(h=\frac{1}{2}gt^{2}\), and \(t = \sqrt{\frac{2h}{g}}\). The time of flight depends only on the height \(h\) and the acceleration due to gravity \(g\). When the height \(h = 5m\) (initial case) and after the initial velocity change (height is still \(h = 5m\)), the time of flight \(T=\sqrt{\frac{2h}{g}}\) remains the same.

Step2: Analyze the horizontal - distance formula

The horizontal distance \(x=v_{0x}t\). Initially, \(D = v_{0x1}T\) (where \(v_{0x1}=20m/s\)). After the initial velocity change, \(v_{0x2} = 40m/s\), and \(t = T\) (from Step 1). The new horizontal distance \(D_{new}=v_{0x2}T\). Substituting \(v_{0x2} = 2v_{0x1}\) (since \(v_{0x1} = 20m/s\) and \(v_{0x2}=40m/s\)) and \(D = v_{0x1}T\) into the formula for \(D_{new}\), we get \(D_{new}=2v_{0x1}T=2D\)

Answer:

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