g force problem?

Question

How fast does a leopard have to be running to reach 12 ft while leaping at a 45 degree angle? HELP! How do I even set this up?

Answer 1

Hi. Assuming 12 feet from take-off to landing, launching at a 45 degree angle cancels most of the trig. Not sure though if you mean 12 height. g is 32'/sec^2, in any case. Since the leopard could, if she wanted to, leap 12 feet straight up then assume the first case is the problem.

Answer 2

Break it into 2 problems.

1. What is the jump velocity required (as if the leopard jumped straight up) to reach 12ft in height. To solve this use the conservation of energy principle. In this case total kinetic energy going up is equal to total potential energy of the leopard being dropped from 12ft.

KE=PE

1/2mass* (velocity)^2 = mass * height* acceleration

Since mass is on both sides it drops out which is good since we don't know how much the leopard weighs. acceleration = 33ft/sec^2

plug and chug and the velocity is equal to 28.14 ft/sec


Now for the angle part...you need a little SOCCAHTOA. In this case draw a triangle. The trajectory of the leopard is the hypoteneuse. The height of the triangle is 28.14 ft/sec. The angle where the leopard leaps up is 45deg.

Sine 45 degrees = opposite/ hypoteneuse.

0.707=(28.14 feet/sec)/hyp

hyp= 39.80 feet/second


The leopard is traveling at 39.8 feet/second

Answer 3

Start at the end of the problem and work backwards.

If you were to fall from a height of 12 ft, how fast would you be going when you hit the ground? This gives you the vertical takeoff velocity that you need to reach the height of 12 feet as well.

Once you have the answer to that, then you need to remember that this velocity is the vertical component of a 45 degree takeoff vector. You'll need the same velocity in the horizontal component of the takeoff vector, and will combine them to get the total speed of the leopard at takeoff.

Answer 4

You first need a basic understanding of vectors and trigonometry.

Let's start with the height. It will determine the speed upward (directly upward in the y-coordinate direction.
12 feet up is 3.66 meters.
Next you need to get the time it's going to take to reach 12 ft up. That will be used to find the upward velocity.
x = 1/2 at^2
3.66 = 1/2 (9.8)t^2
t = 0.864 seconds "rising"
Next the upward velocity.....
v = at
v = (9.8)(0.864) = 8.47m/s

So the upward component has to be 8.47 , the angle at 45 degrees has to be greater than that.
Set up a sin equation
(You need a triangle to show the vectors)
sin = opp/hyp
sin 45 = 8.47 / hyp

hyp = 12 meters/second

OK......for simplicity typing I left out some of the terms in the kinematic equations and the units. Also, you can't see what I did with the vectors. If you need further explanation on this problem, or any other problem, email me at [e-mail]
I can work it out on paper where it's more clear and attach it to an email and send it back to you. That might make more sense.
Have a nice night. - Kevin

Answer 5

Man its been a while, but I think you need two seperate equations for this one. One for the x axis D = V * T. And one for the y axis D = 1/2 g * T^2.

In the y axis it will reach 12', that is d and g is constant 32.2. So T = SQRT(D/(1/2g) = 0.86 seconds. Plug this into the x axis equation and solve for V, D is 12' here too since it is at a 45 degree angle. V = D/T = 12 ft/0.86 s = 13.95 ft/sec.

I believe the answer is 13.95 ft/sec.

Answer 6

This is really pretty easy and doesn't require much math.
A Leopard can jump higher than 12ft without running believe it or not. But jumping up 12ft and out 12ft,(which would be a 45 degree angle)
The hypotenuse of a 12ft by 12ft triangle
a^2 + b^2 = c^2
a=12ft and b=12ft
solve for c
yeap , a LEOPARD could jump that far without a running start!!!!!!!
The answere is Zero.

Answer 7

You don't know the force the leopard pushes off the ground with. Unsolveable.




Silly newtonian physics problems. If its distance you have to assume the motion of the leopard is parabolic and projectile, which is obviously not how things really work. XD Just apply your motion equations.