Question

you are trying to launch a care package to someone who is stranded on the side of a mountain until they can be rescued. Your care package launcher is capable of shooting the package at 89 m/s at 70° above the horizontal. How far away should we place the launcher if the stranded person is 270 m above your location?

Answer 1

If we assume the location of launcher to be origin then equation of package(projectile) are

X equation :

x = u cos$\theta$​​​​​​ t

(since there is no force in x direction, acceleration will be zero in x direction.)

Y equation

y = u sin$\theta$ t + $a_{y}$ $t^{2}$ /2

This time ​​​​​​. Where g is acceleration due to gravity.

ay = -9

If L is position where we should keep the puncher from stranded person then co ordinates of stranded person is

(x, y) = (L, 270)

Substituting these values in X and Y equations:

L = u cos$\theta$ t

270 = u sin$\theta$ t -

gt/2

Eliminating t from these two equations, I.e., substituting t from x equation to y equation :

gL 270 = Ltand - 2ucos20

Given that

= 70°

u = 89 m/sec

g = 9.8 m/$sec^{2}$

We have

270 = 2.747 L - 0.005288 $L^{2}$

Solving this quadratic equation, we get

L = 131.7 m

and L = 387.5 m

Thus launcher can be placed at either of these two distances from stranded person.