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:
[tex]v^2=u^2-2as[/tex]
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:
[tex]\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}[/tex]
Take the square root of both sides:
[tex]\begin{gathered} \sqrt[]{u}^2=\sqrt[]{169.05} \\ \\ u=13.0\text{ m/s} \end{gathered}[/tex]
Therefore, the initial velocity of the ball is 13.0 m/s.
ANSWER:
13.0 m/s
