Question

Students were having a catapult contest, launching a tennis ball with their home built devices. The winner launched a ball at 30 degrees from the ground at 6.0 m/s. How high does the ball go?

Answer 1

The height reached by the ball=0.459 m

Step-by-step explanation:

We are given that

`\theta=30^(\circ)`

Initial speed, u=6 m/s

We have to find the maximum height reached by ball.

Maximum height reached by ball

`h=(u^2sin^2\theta)/(2g)`

Where
`g=9.8m/s^2`

Substitute the values

`h=(6^2sin^230^(\circ))/(2* 9.8)`

`h=(36* (1)/(4))/(2* 9.8)`

`h=0.459 m`

Hence, the height reached by the ball=0.459 m