Question

A football is launched at an angle such that the vertical component of its velocity and the horizontal component of its velocity are both equal to 40 m/s.A. How long does it take for the football to reach its highest point?B. What horizontal distance does the football travel in its time?

Answer 1

A,

The highest point occurs when the vertical velocity is equal to zero.

Then, to find the time needed, we can use the formula below:

$V=V_0+a\cdot t_{}$

Where V is the final velocity after t seconds, V0 is the initial velocity and a is the acceleration.

So, using V = 0, V0 = 40 m/s and a = -10 m/s² (gravity acceleration), we have:

$\begin{gathered} 0=40-10t \\ 10t=40 \\ t=\frac{40}{10} \\ t=4\text{ s} \end{gathered}$

B.

To calculate the horizontal distance traveled in the 4 seconds to reach the maximum height, we can use the formula below:

$d=v\cdot t$

Where d is the distance, v is the horizontal velocity and t is the time.

So, using v = 40 m/s and t = 4 s, we have:

$\begin{gathered} d=40\cdot4 \\ d=160\text{ m} \end{gathered}$

Also, to find the total horizontal distance traveled until the football reaches the ground again, we can use the time of flight as double the time to reach the maximum height:

$\begin{gathered} t=2\cdot4=8\text{ s} \\ d=40\cdot8 \\ d=320\text{ m} \end{gathered}$