Projectile motion without Vi or time known

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jennknoe
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1.An airplane is trying to deliver medical supplies by dropping them from the cargo bay during a fly-over mission (they have no initial velocity RELATIVE to the plane). Due to mountains in the area, when the plane drops the supplies, it is moving at an angle of 15o above horizontal, and it is at a height of 100m above the target.

a) Find the speed of the airplane needs to have in order to hit the target if the horizontal distance to the target is 1km.

b)Find the velocity of the package of medical supplies just before they hit the ground.

Homework Equations


Xf = X0 + Vi + 1/2at2


The Attempt at a Solution


So to be honest, I'm definitely not that good at physics, and this course is really difficult for me. So sorry if I ask a lot of questions and don't help out too much.

I tried first trying to find time it takes the package to get reach the ground. But I wasn't sure how to find this with no initial velocity with the plane. Then I also noticed that I had to find the time taken for it to travel on the 15 degrees down to horizontal, then how long it would take to fall the rest of the 100m. Any help would be great!

Thanks everyone,
Jenny
 
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I assume "dropping" means they impart no velocity to the package relative to the plan. That is, they literally let go of the package and let it fall. That means its initial velocity is exactly the same as the plane's velocity.

From there you can find the next part (hint: there's nothing special about falling back to the drop altitude -- keep going).
 
jennknoe said:

Homework Equations


Xf = X0 + Vi + 1/2at2

First of all your equation is not quite right. It should be

[tex]x_f = x_o + v_i t + \frac{1}{2} a t^2[/tex]

You just forgot the "t" in the second term on the right side of the equation.

Also notice that there is only acceleration in the y dimension. So your x dimension equation should be

[tex]x = x_0 + v_i cos(15) t[/tex]

and since your y dimension is accelerated, your equation in the y dimension is

[tex]y = y_0 + v_i sin (15) t - \frac{1}{2} g t^2[/tex]

where g is a positive quantity.

Make sure that you understand where these equations come from, then try using them and see where you get.