Q2.2 Vertical motion under gravity (S)
A ball is thrown upwards with speed 4 ms-1 from a height of 1 m above the ground. At what speed will it hit the ground?
The same ball is thrown downwards with speed 4 ms-1 from the same height. At what speed will it now hit the ground?
Draw graphs of (vertical) velocity against time to illustrate both cases.
Solution
Reveal

Solution
Denote the launch height by h. Using the third of the constant acceleration equations Key Point 1.4 applied to the vertical (y) motion we have
![\[ v^2=v_{0}^2 +2 a(y-y_0)=v_{0}^2 -2g(y-y_0) \]](mastermathpng-0.png)
When the downwardly thrown ball reaches the ground its y displacement is
![\[ v^2=v_{0}^2 +2gh \]](mastermathpng-1.png)
![\[ v^2=16+20=36 \hspace{0.5cm}\mbox{\rm implying}\hspace{0.5cm} v=\pm 6\, ms^{-1} \]](mastermathpng-2.png)
The solution we want is v=-6 ms-1 signalling downward motion.
Aside: The positive solution simply tells us the upward velocity with which the particle would have had to be launched from the ground in order to pass through the point y=1 m with speed 4 ms-1.
If the ball is launched upwards rather than downwards we just
change the sign of v0. Since the equation we have just solved involves the
square of v0 the result for the final velocity is unchanged. The v-t graphs are shown. In each case the slope is -g. |