Projectile motion from an angled slope

Click For Summary
To determine the optimal launch angle for a rock thrown from a slope at 30 degrees, the projectile motion equations must be adapted to a new coordinate system where the slope is the x-axis. The key is to express the range in terms of the launch angle and then maximize this expression. It is noted that the optimal angle will not be 45 degrees due to the slope's inclination. A diagram can aid in visualizing the motion, and solving for the angle relative to the x-axis simplifies the calculations. Ultimately, the goal is to find the angle that maximizes the range when the projectile lands.
BarneyStinson
Messages
10
Reaction score
0

Homework Statement



A rock is thrown from the top of a slope that makes an angle of 30 degrees with the horizontal. At what angle to the horizontal should the rock be thrown to get a maximum range? (Hint: pick the direction of the slope as a new x-axis and the normal to the slope as the new y axis. There will be acceleration in both axes.)

Homework Equations



y=y_o+v_y_o*-.5*g*t^2
v_y=v_y_o+a_y*t

x=x_o+v_x_o+.5a*t^2
v_x=v_x_o

The Attempt at a Solution



I drew a diagram of what is happening:

2wddod1.jpg


I need to solve for theta.

I don't think it will be 45 degrees to the horizontal, as you start on a slant. Even if i did think it was, i don't know where to plug 45 into.

Can anyone guide me on the right path? Offer any more hints?

Thanks.
 
Physics news on Phys.org
You'll have to write the equations of motion for the projectile in the given coordinate system and find an expression for the range (the x coordinate when the y coordinate reaches zero --- the projectile lands). The trick will be in maximizing this expression with respect to the launch angle. I think it would be easier to first solve for the angle with respect to the x-axis. It can always be adjusted to the horizontal reference afterwards.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 36 ·
2
Replies
36
Views
4K
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K