Question

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?

Answer 1

The ball's position vector has components

`x=\left(8.00(\rm m)/(\rm s)\right)\cos40.0^\circ t`

`y=1.00\,\mathrm m+\left(8.00(\rm m)/(\rm s)\right)\sin40.0^\circ t-\frac g2t^2`

where
`g=9.80(\rm m)/(\mathrm s^2)` is the acceleration due to gravity. The ball hits the ground when
`y=0`:

`0=1.00\,\mathrm m+\left(8.00(\rm m)/(\rm s)\right)\sin40.0^\circ t-\frac g2t^2\implies t=1.22\,\mathrm s`

The ball's velocity vector has components

`v_x=\left(8.00(\rm m)/(\rm s)\right)\cos40.0^\circ`

`v_y=\left(8.00(\rm m)/(\rm s)\right)\sin40.0^\circ-gt`

so that after 1.22 s, the velocity vector is

`\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)(\rm m)/(\rm s)`

and the magnitude is

`\|\vec v\|=√(6.13^2+(-6.79)^2)\,(\rm m)/(\rm s)=\boxed{9.14(\rm m)/(\rm s)}`