Question

How do i solve this problem? Hint: The cannonball is being launched vertically upwards, therefore, there is no initial horizontal speed. The given initial speed will also be the initial vertical speed.

Answer 1

The initial velocity of the ball is given as 36.0 m/s.

The horizontal component of velocity of ball is given as,

`v_x=v\cos \theta`

The ball is projected vertically, therefore, the angle made by ball is 90 degree.

Plug in the known values,

`\begin{gathered} v_x=(36.0m/s)cos90^(\circ) \\ =(36.0\text{ m/s)(0)} \\ =0\text{ m/s} \end{gathered}`

Therefore, the initial horizontal velocity of ball is 0 m/s.

The vertical component of velocity of ball is given as,

`v_y=v\sin \theta`

Plug in the known values,

`\begin{gathered} v_y=(36.0m/s)\sin 90^(\circ) \\ =(36.0\text{ m/s)(1)} \\ =36.0\text{ m/s} \end{gathered}`

Therefore, the initial vertical velocity of the ball is 36.0 m/s.