Question

An astronaut on the Moon, where there is no air resistance, throws a ball. The ball’s initial velocity has a vertical component of 8.00 ms–1 and a horizontal component of 4.00 m s–1, as shown. The acceleration of free fall on the Moon is 1.62m s–2. What will be the speed of the ball 9.00 s after being thrown? A 6.6 m s–1 B 7.7 m s–1 C 10.6 m s–1 D 14.6 m s

Answer 1

As the horizontal component remains the same throughout the motion we have to find the vertical component at 9.0 seconds. Use v=u+at to find vertical component and then use pythagores theorem to calculate the reaultant velocity at 9.0 seconda.
Thats all.

Answer 2

The speed of the ball is $\bold{7.7 \ \mathrm{ms}^{-1}}$

Explanation:

Let the vertical component of the velocity be its initial vertical velocity denoted as $V_1$ and the horizontal component is a constant velocity and it is denoted as $V_H$. The final velocity after 9 s is the vector sum of the final vertical velocity denoted as $V_2$ and the horizontal velocity $V_H$.

It is known that,

$\mathrm{V}_{1}=8 \ \mathrm{ms}^{-1}$

$\mathrm{V}_{\mathrm{H}}=4 \ \mathrm{ms}^{-1}$

$g=1.62 \ \mathrm{ms}^{-2}$

From the equations of motion of free fall,

$V_{1}=g t$

$\Text{Time \ taken \ to \ reach \ the \ maximum \ altitude, \ t} = \frac{V_{1}}{g}$

$t=\frac{8}{1.62}=4.94 \ s$

So the maximum altitude is reached at 4.94 s. So the time taken by the ball to fall from the maximum altitude to the position at 9 s is,

$t^{\prime}=9-4.94=4.06 \ s$

Now, the velocity of the ball during falling from the maximum altitude to the position at 9 s is,

$V_{2}=g t^{\prime}$

$V_{2}=1.62 \times 4.06=6.58 \ \mathrm{m} / \mathrm{s}$

The total final velocity at 9 s,

$V^{2}=V_{H}^{2}+V_{2}^{2}$

$V^{2}=V_{H}^{2}+V_{2}^{2}$

$V^{2}=59.30$

On taking square root, we get,

$V=7.7 \ \mathrm{m} / \mathrm{s}$