Question

A secret agent skis off a slope inclined at θ = 30.2 degrees below horizontal at a speed of v0 = 20.4 m/s. He must clear a gorge, and the slope on the other side of the gorge is h = 11.7 m below the edge of the upper slope. Does he make it?

Answer 1

Answer:

He will make it if the gorge is no wider than 14.4 m

Explanation:

The secret agent follows a parabolic motion. We have the following data:

$v_0 = 20.4 m/s$ is the initial speed

$\theta=30.2^{\circ}$ below the horizontal is the initial angle

h = 11.7 m is the vertical distance covered by the agent before landing on the other side

Let's start by analyzing the vertical motion. The initial vertical velocity is

$u_y = v_0 sin \theta = (20.4) (sin 30.2^{\circ})=10.3 m/s$

Where we have chosen downward as positive direction. Now we use the following equation:

$h=u_y t + \frac{1}{2}gt^2$

where $g=9.8 m/s^2$ (acceleration of gravity) to find the time t at which the agent lands. Substituting the numbers:

$11.7 = 10.3 t + 4.9 t^2\\4.9t^2 + 10.3t -11.7 = 0$

Which has two solutions: t = -2.92 s and t = 0.82 s. Since the negative solution is meaningless, we discard it, so the agent reaches the other side of the gorge after 0.82 s.

Now we want to find what is the maximum width of the gorge that allows the agent to safely land on the other side. For that, we need to calculate the horizontal velocity of the agent, which is constant during the motion:

$u_x = u_0 cos \theta = (20.4)(cos 30.2^{\circ})=17.6 m/s$

So, the horizontal distance covered by the agent is

$d = u_x t = (17.6)(0.82)=14.4 m$

So, the agent will land safely if the gorge is at most 14.4 m wide.