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1. Mar 25, 2016

RoboNerd

1. The problem statement, all variables and given/known data
An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for he object to travel a maximum horizontal distance from the cliff before hitting the ground, the throw should be at an angle theta with respect to the horizontal of

A) greater than 60 degrees above the horizontal
B) greater than 45 degrees but less than 60 degrees above the horizontal
C) greater than zero but less than 45 degrees above the horizontal
D) zero
E) greater than zero but less than 45 degrees below the horizontal.

2. Relevant equations
Kinematics. range = Vx * flight time

3. The attempt at a solution

I put down D with the following logic: if I have my velocity at an angle, then its horizontal component would be smaller than if the velocity is directed along an angle of theta=0. Increased horizontal component would increase its range per the equation that I typed above.

The solutions say that the answer is C. Why is this the case, and why are each of the answers that are not C incorrect?

Thanks in advance for the help!

2. Mar 25, 2016

drvrm

this logic is not correct -as when you throw with an angle say theta with the horizontal

your projectile returns to level of the cliff covering the range (R1) by its horizontal component
but again it is going at an angle towards the ground and

covers additional horizontal path(R2) and one should maximize the
total path( R1+ R2 ) which will be necessarily larger than your logic- range.
try to see at what angle the range is maximum.

3. Mar 25, 2016

RoboNerd

it is 45 degrees that will have maximum range, right?

4. Mar 25, 2016

PeroK

You can sort of work this out by thinking about it, but to resolve the question definitely you need to use ... wait for it ... mathematics!

5. Mar 26, 2016

ehild

No, it is true only when the object returns at the same height.
Imagine a stone projected from the edge of a 100 m high cliff. When does it travel a longer horizontal distance, when projected horizontally or at 45°?

6. Mar 26, 2016

drvrm

write out the formula for the range (total ) and try to maximize.

7. Mar 26, 2016

PeroK

That might not be so easy, actually!

8. Mar 26, 2016

ehild

Just some more help for the OP or anyone, interested in the problem: The formula for the trajectory is known. If h is the height of the cliff and the y coordinate is zero on the ground,
y=h+xtan(θ)-0.5x2/(v0cos(θ))2. We get the range X when y=0.
h+xtan(θ)-0.5x2/(v0cos(θ))2=0 *
Apply implicit differentiation: the range can be maximum when dx/dt=0. This condition gives a relation between the range X and θ. Substituting back to equation * you get θ in terms of h, and you can decide, if θ is greater or smaller than 45° if h>0.

9. Mar 26, 2016

PeroK

Just a minor point that $g$ is missing from that equation:

$y=h+xtan(θ)-0.5gx^2/(v_0cos(θ))^2$

10. Mar 26, 2016

ehild

Oh, sorry, Thank you!

11. Mar 27, 2016

RoboNerd

How did you derive the tan term in the equation?

Thanks.

12. Mar 27, 2016

PeroK

It's probably not how you were supposed to solve this problem, but it's a good exercise to derive that parabolic formula relating $x$ and $y$. Hint: express $x$ and $y$ in terms of $\theta$ and $t$ and use substitiution to eliminate $t$.

13. Mar 28, 2016

RoboNerd

Good point. I just derived the formula.

However.... this question comes from an ap physics exam multiple choice section, and your approach with deriving the equation and doing implicit differentiation just seems too time consuming for the test.

What is the simplest method for solving it that you can think of?

14. Mar 29, 2016

ehild

The formula y(x) is usually taught when studying projectile motion.
I think, the method I suggested is the simplest solution, and it would be useful for you to try it. But you can guess the correct choice if you read my post #5 again. 45°is the optimum angle if there is no height difference between the initial and final points of the trajectory. You get the range if you multiply the horizontal component of the initial velocity by the time of flight. That time is the sum of the time of rise and the time of fall. Assume that the cliff is very high. Then the time of rise can be much less than the time of fall. When do you get a longer range, if you project the object at higher angle or smaller angle than 45°?

15. Mar 29, 2016

atom jana

Yeah, 45 degrees above the horizontal should have the maximum range.

16. Mar 29, 2016

PeroK

I would reason as follows:

1) It can't be > 45° because when it reaches the height of the cliff again, it is not as far and travelling slower in the x-direction than a 45° trajectory. So 45° is always better than > 45°.

2) It can't be < 0° because that's obviously worse than a trajectory of 0°.

So, the optimum must be somewhere between 0° and 45° (above the horizontal).

A bit more thought leads to:

3) If the cliff is low, the optimum trajectory will be close to 45°; if the cliff is very high, the optimum trajectory will be close to 0°.

4) I wonder whether the optimum angle for a cliff of a given height depends on the velocity? Perhaps the faster you throw it, the lower the optimum angle?

This last point, however, is where probably you need some maths to back up and confirm this intuitive reasoning.

17. Mar 29, 2016

ehild

The following equation holds for the angle at maximum range: $$\sin^2(\theta)= \frac{0.5}{\frac{hg}{v_0^2}+1}$$

18. Mar 29, 2016

RoboNerd

How did you derive this relationship? Is this some variation of the range formula?

Apart from that, I understand your approaches. Thanks so much for the help!

19. Mar 29, 2016

ehild

I did the implicit differentiation of the y(x )=0 formula with respect to theta. I got a relationship between the range x and theta. I replaced it back into the equation y(x)=0, and isolated sin2(θ).

20. Mar 29, 2016

RoboNerd

Sounds like an overcomplicated approach IMO