Question

A student fires a cannonball vertically upwards. The cannonball returns to the ground after a 3.80s flight. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. Use regular metric units (ie. meters).How long did the cannonball rise? What was the cannonball's initial speed? What was the cannonball's maximum height?

Answer 1

The time taken by ball to rise is given as,

`t=3.80\text{ s}`

Therefore, the total time taken to move the ball in the upward direction is calculated as,

`\begin{gathered} t_0=\frac{3.80\text{ s}}{2} \\ =1.90\text{ s} \end{gathered}`

Therefore, the ball rise for 1.90 s.

The final speed of the ball can be given as,

`v=u+gt`

At the maximum height the final speed of ball is zero.

Substitute the known values,

`\begin{gathered} 0m/s=u+(9.8m/s^2)(1.90\text{ s)} \\ u=-(9.8m/s^2)(1.90\text{ s)} \\ =-18.62\text{ m/s} \end{gathered}`

Therefore, the initial speed of the ball is -18.62 m/s where negative sign indicates the direction of ball.

The final speed of the ball can be given as,

`v^2=u^2-2gh_m`

At the maximum height the final speed is zero. Substitute the known values,

`\begin{gathered} (0m/s)^2=(-18.62m/s)^2-2(9.8m/s^2)h_m \\ h_m=((-18.62m/s)^2)/(2(9.8m/s)^2) \\ =17.7\text{ m} \end{gathered}`

Therefore, the maximum height of the canon is 17.7 m.