(2D problem) What angle will yield maximum horizontal range?

[physics1829]

Homework Statement

A supply plane is flying at 60.0m/s at an altitude of 200.0m. What angle of flight will yield the maximum horizontal range for the dropped package?

Homework Equations

1/2(g)(t)².

The Attempt at a Solution

So far the only equations that i have, but may be wrong, is x=((60m/s)cos([itex]\Theta[/itex]))t+1/2(g)(t)². and y=((60m/s)sin([itex]\Theta[/itex]))t+1/2(g)(t)²+200m. I'm told it would involve a quadratic formula if i do it the more efficient way, but i cannot figure out how to complete it, or if i am going about it all wrong to find the angle that gives the farthest distance. I have also tried messing around with the formulas to try to combine them. I've also thought of using excel to test every angle from 0.0-90.0° but i haven't worked out how i would do that either.

 

[tiny-tim]

Welcome to PF!

Hi Wily Fury! Welcome to PF!

x=((60m/s)cos([itex]\Theta[/itex]))t+1/2(g)(t)²

nooo …

in the x direction, a = … ? ​

 

[physics1829]

the acceleration should be 0 for the x direction?

 

[tiny-tim]

Yup!

 

[physics1829]

i'm still not sure how to approach the next step or if i am on the right track.

 

[tiny-tim]

i'm still not sure how to approach the next step or if i am on the right track.

The next step is to get an expression for the range (as a function of θ).

Then find the maximum value of the range, either by differentiating wrt θ, or by completing the square.

(and now I'm off to bed :zzz:)

 

[physics1829]

I'm not really sure what that means >.<

 

[tiny-tim]

what what means?

 

[haruspex]

i'm still not sure how to approach the next step or if i am on the right track.

You have (correcting your sign for vertical acceleration) x = u cos(θ)t, y = h + u sin(θ)t - gt²/2. Try to answer the following in sequence:
- at what time (as a function of h, u, g, theta) will the package hit the ground?
- what will x be at that time?
- how do you maximise that wrt theta?