2021 A/L Physics Past Paper · Question 48 (English Medium)

Answer: (3) $30\ \text{m s}^{-1}$

Given information

We know:

The ball is thrown up from the ground.

It passes the height $h = 25\ \text{m}$ twice, $4\ \text{s}$ apart.

Air resistance is neglected, so $g = 10\ \text{m s}^{-2}$.

What the question asks

We want: the initial velocity $u$.

Relevant theory

The motion under the action of gravitational force, with negligible resistances, such as air resistance, is called motion under gravity.

$$s = \left(\frac{u + v}{2}\right)\left(\frac{v - u}{a}\right) \Rightarrow v^2 = u^2 + 2as$$

The ball goes up past $25\ \text{m}$, stops at the top, and comes down past $25\ \text{m}$ again.

The way down is the way up played backwards.

So the $4\ \text{s}$ is split into two equal halves.

Required equations

$t = \dfrac{4}{2}$

$0 = v_{25} - g t$

$v_{25}^2 = u^2 - 2 g h$

Applying it to this question

All the units are in m and s.

So the answer comes out in $\text{m s}^{-1}$.

$2$

$t = \dfrac{4}{2} = 2$

$20$

$v_{25} = g t = 10 \times 2$

$20^2 = u^2 - 2 \times 10 \times 25$

$u^2 = 400 + 500 = 900$

$30$

$u = 30$

Reasoning and calculation

The ball needs $2\ \text{s}$ to go from $25\ \text{m}$ to the top.

So its speed at $25\ \text{m}$ is $g \times 2 = 20\ \text{m s}^{-1}$.

It was faster at the ground, because it slowed down while climbing $25\ \text{m}$.

That gives $30\ \text{m s}^{-1}$.

Option (3).

Checks

$sqrt(v^2 + 2 * g * h)$

Does it make sense?

The height of the top is:

$$H = \dfrac{u^2}{2g} = \dfrac{900}{20} = 45\ \text{m}$$

That is above $25\ \text{m}$, as it should be.

The ball takes $\dfrac{30}{10} = 3\ \text{s}$ to reach the top.

At $2\ \text{s}$ before the top it is at $45 - \tfrac{1}{2} \times 10 \times 2^2 = 25\ \text{m}$. This agrees.

Conclusion

$u = \sqrt{20^2 + 2 \times 10 \times 25} = 30\ \text{m s}^{-1}$

The initial velocity is $u = 30\ \text{m s}^{-1}$.

The answer is (3).

A ball thrown up passes every height twice, once up and once down.

Half of the time between the two passes is the time to the top.

Use that half to find the speed at that height.

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