Vertical Motion due to Gravity

A ball \(\mathbf{A}\) falls vertically from rest from the top of a tower 63 m high. At the same time as \(\mathbf{A}\) begins to fall, another ball \(\mathbf{B}\) is projected vertically upwards from the bottom of the tower with speed 21 ms\(^{-1}\). The balls collide. Find the distance of the point where the balls collide from the bottom of the tower.

- Motion for \(\mathbf{A}\)
- \(u = 0\)
- \(a=9.8\)
- \(s=ut+\frac{1}{2}at^2\) (because we cant find \(t\) right now we express in terms of \(t\))
- \(s_{1}=4.9t^2\)

- Motion for \(\mathbf{B}\)
- \(u=21\)
- \(a=-9.8\)
- \(s=ut+\frac{1}{2}at^2\) (because we cant find \(t\) right now we express in terms of \(t\))
- \(s_{2}=21t-4.9t^2\)

- \(s_{\text{total}}=63\)
- \(63=4.9t^2-4.9t^2+21t\)
- \(21t=63\)
- So it takes the two balls 3 seconds to collide.

- Find distance FROM BOTTOM of the tower
- \(s_{2}=21t-4.9t^2\)
- \(s_{2}=21\times3-4.9\times9\)
- \(s_{2}=18.9\)