Question 1.8: A juggler throws a ball up in the air with initial speed 5 m...
A juggler throws a ball up in the air with initial speed 5 ms^{-1} from a height of 1.2 m. It has a constant acceleration of 10 ms^{-2} vertically downwards due to gravity.
i) Find the maximum height of the ball above the ground and the time it takes to reach it.
At the instant that the ball reaches its maximum height, the juggler throws up another ball with the same speed and from the same height.
ii) Where and when will the balls pass each other?
i) In this example it is very important to draw a diagram and to be clear about the position of the origin. When O is 1.2 m above the ground and s is the height in metres above O after t s, the diagram looks like figure 1.34.
At the point of maximum height,
The ball stops instantaneously before falling so at the top v = 0.
An equation involving u, v, a and s is required.
v² = u² + 2as
\begin{matrix}0 = 5² + 2 × (-10) × H &\longleftarrow \end{matrix} \begin{matrix} \boxed{\text{The acceleration given is constant, } a = -10; \ u = +5; \ v = 0 \text{and} s=H.}\\ \end{matrix}
H = 1.25
The maximum height of the ball above the ground is 1.25 + 1.2 = 2.45 m.
To find t_{1}, given v = 0, a = -10 and u = + 5 requires a formula in v, u, a and t.
v = u + at
0 = 5 + (-10) t_{1}
t_{1} = 0.5
The ball takes half a second to reach its maximum height.
ii) Now consider the motion from the instant the first ball reaches the top of its path and the second is thrown up.
Suppose that the balls have displacements above the origin of x_{1} m and x_{2} m, as shown in the diagram, at a general time t s after the second ball is thrown up. The initial position of the second ball is zero, but the initial position of the first ball is +1.25 m.
For each ball you know u and a. You want to involve t and s so you use
s – s_{0} = ut + \frac{1}{2}at^{2}i.e. s = s_{0} + ut + \frac{1}{2}at^{2}
For the first ball:
\begin{matrix}x_{1} = 1.25 + 0 × t + \frac{1}{2} × (-10) × t^{2} &\longleftarrow \end{matrix} \begin{matrix} \boxed{\text{This makes } x_{1} = 1.25 \text{ when} t = 0.}\\ \end{matrix}
\begin{matrix}x_{1} = 1.25 – 5t^{2} ① &\longleftarrow \end{matrix} \begin{matrix} \boxed{x_{1} \text{decreases as t increases. }} \\ \end{matrix}
For the second ball:
x_{2} = 0 + 5 × t + \frac{1}{2} × (-10) × t^{2}
x_{2} = 5t – 5t^{2} ②
Suppose the balls pass after a time t s. This is when they are at the same height, so equate x_{1} and x_{2} from equations ① and ②
1.25 – 5t² = 5t – 5t²
1.25 = 5t
t = 0.25
Then substituting t = 0.25 in ① and ② gives
\begin{matrix} x_{1} = 1.25 – 5 × 0.25² = 0.9375 \longleftarrow & \boxed{\text{These are the same, as expected.} } \\and \\ x_{2} = 5 × 0.25 – 5 × 0.25² = 0.9375 . \swarrow \end{matrix}
The balls pass after 0.25 seconds at a height of 1.2 + 0.94 m = 2.14 m above the ground (correct to the nearest centimetre).
Note
The balls pass after half the time to reach the top, but not half-way up.
