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?

Question

A juggler throws a ball up in the air with initial speed 5 [latex]ms^{-1}[/latex] from a height of 1.2 m. It has a constant acceleration of 10 [latex]ms^{-2}[/latex] 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?

Accepted Answer

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,

[latex]\begin{matrix} ext{ let } s = H ext {and} t = t_{1}. &\longleftarrow \end{matrix} \begin{matrix} \boxed{ ext{Use the suffix because there are two times to be found in this question.}}\ \end{matrix}[/latex]

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

[latex]\begin{matrix}0 = 5² + 2 × (-10) × H &\longleftarrow \end{matrix} \begin{matrix} \boxed{ ext{The acceleration given is constant, } a = -10; \ u = +5; \ v = 0 ext{and} s=H.}\ \end{matrix}[/latex]

H = 1.25

The maximum height of the ball above the ground is 1.25 + 1.2 = 2.45 m.

To find [latex]t_{1}[/latex], given v = 0, a = -10 and u = + 5 requires a formula in v, u, a and t.

v = u + at

0 = 5 + (-10) [latex]t_{1}[/latex]

[latex]t_{1}[/latex] = 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 [latex]x_{1}[/latex] m and [latex]x_{2}[/latex] 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

[latex]s - s_{0} = ut + \frac{1}{2}at^{2}[/latex]

i.e. [latex]s = s_{0} + ut + \frac{1}{2}at^{2}[/latex]

For the first ball:

[latex]\begin{matrix}x_{1} = 1.25 + 0 × t + \frac{1}{2} × (-10) × t^{2} &\longleftarrow \end{matrix} \begin{matrix} \boxed{ ext{This makes } x_{1} = 1.25 ext{ when} t = 0.}\ \end{matrix}[/latex]

[latex]\begin{matrix}x_{1} = 1.25 - 5t^{2} ① &\longleftarrow \end{matrix} \begin{matrix} \boxed{x_{1} ext{decreases as t increases. }} \ \end{matrix}[/latex]

For the second ball:

[latex]x_{2} = 0 + 5 × t + \frac{1}{2} × (-10) × t^{2}[/latex]

[latex]x_{2} = 5t - 5t^{2}[/latex] ②

Suppose the balls pass after a time t s. This is when they are at the same height, so equate [latex]x_{1}[/latex] and [latex]x_{2}[/latex] 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

[latex]\begin{matrix} x_{1} = 1.25 - 5 × 0.25² = 0.9375 \longleftarrow & \boxed{ ext{These are the same, as expected.} } \and \ x_{2} = 5 × 0.25 - 5 × 0.25² = 0.9375 . \swarrow \end{matrix} [/latex]

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.

Question 1.8: A juggler throws a ball up in the air with initial speed 5 m...

Related Answered Questions

Verified Answer:

Verified Answer:

Verified Answer:

A car moves between two sets of traffic lights, stopping at both. Its speed v ms^-1 at time t s is modelled by v = 1/20 t (40 – t), 0≤t≤40 . Find the times at which the car is stationary and the distance between the two sets of traffic lights. ...

Verified Answer:

The velocity (in ms^-1) of a model train which is moving along straight rails is v = 0.3t² – 0.5 Find its displacement from its initial position i) after time t ii) after 3 seconds. ...

Verified Answer:

Verified Answer:

David walks east for 6 s at 2 ms^-1 then west for 2s at 1 ms^-1. Draw i) a diagram of the journey ii) the speed–time graph iii) the velocity–time graph. Interpret the area under each graph. ...

Verified Answer:

A coin is dropped from rest at the top of a building of height 12 m and travels in a straight line with constant acceleration 10 ms^-2. Find the time it takes to reach the ground and the speed of impact. ...

Verified Answer:

When leaving a town, a car accelerates from 30 mph to 60 mph in 5 s. Assuming the acceleration is constant, find the distance travelled in this time. ...

Verified Answer:

Verified Answer: