Question

A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when it is at an altitude of 600 m. If air resistance is ignored, the horizontal distance of the point of impact from the plane at the moment of the package's release is what? 1. 720 m.
2. 420 m.3. 2800 m.
4. 6800 m
5. 5500 m

Answer 1

Answer:

The correct option is 1.  720 m

Explanation:

Projectile Motion

When an object is launched in free air (no friction) with an initial speed vo at an angle $\theta$, it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.

$\displaystyle V_o=250\ m/s$

$\displaystyle \theta =-37^o$

The initial velocity has these components in the x and y coordinates respectively:

$\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s$

$\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s$

And we know the plane has an altitude of 600 m, so the package will reach ground level when:

$\displaystyle y=-600\ m$

The vertical distance traveled is given by:

$\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600$

We'll set up an equation to find the time when the package lands

$\displaystyle -150.5t-4.9\ t^2=-600$

$\displaystyle -4.9\ t^2-150.5\ t+600=0$

Solving for t, we find only one positive solution:

$\displaystyle t=3.6\ sec$

The horizontal distance is:

$\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m$

The correct option is 1.  720 m