Question

Could I get help with number 2 in the following image

Answer 1

Given:

Height of ramp = 19.6 m

Final Velocity = 4.9 m/s

The angle slows down the ball by 3.7 m/s

Let's find the initial velocity of the ball.

To find the initial velocity, apply the motion formula:

`v^2=u^2-2as`

Where:

v is the final velocity = 4.9 m/s

u is the intial velocity

Since the ball slows by 3.7 m/s each second, the decceleration, a = 3.7 m/s²

s is the height = 19.6 m

Rewrite the formula for v and input values into the formula:

`\begin{gathered} u^2=v^2+2as \\ \\ u^2=4.9^2+2(3.7)(19.6) \\ \\ u^2=24.01+145.04 \\ \\ u^2=169.05 \end{gathered}`

Take the square root of both sides:

`\begin{gathered} \sqrt[]{u}^2=\sqrt[]{169.05} \\ \\ u=13.0\text{ m/s} \end{gathered}`

Therefore, the initial velocity of the ball is 13.0 m/s.

ANSWER:

13.0 m/s