Question

A tennis ball “A” is released from rest down a 10.0 m long inclined ramp with a uniform acceleration of 5.0 m/s2. Another tennis ball “B” is initially located at the same height as ball “A” right above the lower edge of the ramp. Ball “B” is thrown upward with some initial speed at the same instant as the release of ball “A”.
a) What was the initial velocity of ball "B" so that "A" and "B" reach the bottom of the ramp at the same time?

Answer 1

Answer:

Please find the complete question in the attached file.

Explanation:

The time is taken by ball A to reach the bottom $\to t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2 \times 10}{5}}=\sqrt{\frac{2 0}{5}}= \sqrt{4}=2\ s$

Calculating the velocity of the ball:

$-h=ut-\frac{1}{2}gt^2\\\\-S \sin 30=ut-\frac{1}{2}gt^2\\\\-5=u(2)-\frac{1}{2}\times 9.81 \times 2^2$

$-5=2u-9.81 \times 2\\\\-5=2u-19.62\\\\-5+19.62=2u\\\\14.62=2u\\\\u=\frac{14.62}{2}\\\\u=7.31 \ \frac{m}{s}$