Question

A $55kg$ astronaut who weighs $180N$ on a distant planet is pondering whether she can leap over a$3.5$-m-wide chasm without falling in. If she leaps at a $15°$ angle, what initial speed does she need to clear the chasm?

Short answer

The astronaut should launch at initial velocity $v_o=4.8\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given Information

A $55kg$ astronaut who weighs $180N$ on a distant planet is pondering.

Explanation

We use coordinates with the $x$- axis horizontal and the $y$-axis vertical. To pass the chasm, the astronaut must use the projectile motion where projectile motion is motion under the influence of only gravity . So, we must calculate the gravitational acceleration $g$on that planet. We are given the Mass and the Weight of the astronaut where the acceleration is constant in the vertical direction with $a_y=g$. Newton 's second law states that the weight equals the mass time the gravitational acceleration

$w=mg$

Solve equation (1) for $g$and plug the values for $m$ and $w$ to get $g$

$g=\frac{w}{m}=\frac{180\mathrm{N}}{55\mathrm{kg}}=3.27\mathrm{m}/{\mathrm{s}}^2$

Explanation

The astronaut model as a particle launched with speed $v_0$at an Angle $\theta$. The distance that the astronaut travels is called its range and it in this problem equals the width of the chasm$d$. In our problem, the astronaut lands at the same level from which she was launched, and because of that we can use equation (4.15) in situations like this, where equation (4.15) states the range of the projectile by

$\text{Range}=d=\frac{v_o^2\sin \left(2\theta \right)}{g}$

Where $\theta$is the angle of launch, $v_o$is the initial velocity and the range will be the width of the chasm $d$. Solve equation (2) for $v_o$

$d=\frac{v_o^2\sin \left(2\theta \right)}{g}$

$v_o=\sqrt{\frac{dg}{\sin \left(2\theta \right)}}$

Explanation

Now, plug the values for $\theta ,d$and $g$into equation (3) to get $v_o$

$v_o=\sqrt{\frac{dg}{\sin \left(2\theta \right)}}$

$=\sqrt{\frac{(3.5\mathrm{m})\left(3.27\mathrm{m}/{\mathrm{s}}^2\right)}{\sin \left(2\left[15°\right]\right)}}$

$=4.8m/s$

Final Answer

The astronaut should launch at initial velocity $v_o=4.8\mathrm{m}/\mathrm{s}$.