Question

A 200 lb. circus performer jumps from a 20 foot ladder onto a teeter-totter, launching a 50 lb. child. Assume a perfect transfer of energy, with no friction or air resistance. In his arc of travel, the child lands 50 feet horizontally to the rear. What was the total height the child achieved?

Answer 1

not sure how to do the work to get the answer but my brother came over and told me the answer but idk how he got it....

Answer 2

I know the following: \[E _{T} = E _{P} + E _{K}\]\[E _{P} = wh\]\[E _{K} = 0.5mV ^{2}\]From what I calculated, the total energy equals 4,000 ft-lbs, and since we're assuming a perfect transfer of energy, the child is also going to feel 4,000 ft-lbs of energy. Therefore, the child should achieve a height of 80 feet going straight up. However, I'm not sure how to take the 50 feet of horizontal travel into consideration and adjust the actual total height accordingly.

Answer 3

do you have answer choices.

Answer 4

@nathanruff

Answer 5

According to my instructor, the answer is 75.9 feet.

Answer 6

thats exactly the answer

Answer 7

Could you please explain how you got 75.9 feet then?

Answer 8

like i said before i have no idea how my brother got that he just told me the answer

Answer 9

I just glanced at the problem... so I'm not entirely sure... But my guess would be to start by saying that the initial total energy (which is the potential energy of the 200lb person), equals the kinetic energy of the of the child as soon as they are launched. You can use that to find the launch speed of the child. Then if you know the initial speed, the acceleration of gravity, and the final horizontal distance, I'd assume you'd be able to solve for the max height

Answer 10

right so once you find the initial speed of the child (v_i), you can say that
v_i^2 = v_x^2 + v_y^2
horizontal velocity is constant so you can say that time is the horizontal distance traveled divided by the horizontal velocity
t = x / v_x
using geometry you can say that the launch angle (theta) is related to the direction of the velocity
tan(theta) = v_y / v_x
theta = arctan(v_y / v_x)

Now we can bring in the projectile equations:
Time of flight:
t = 2*v_i *sin(theta) / g
Max Height:
H = v_i^2 * sin^2(theta) / (2*g)
Horizontal Range:
x = v_i^2 * sin(2*theta) / g

so we now have 3 equations (where you can plug in your expression for theta and time) and 3 unknowns (height, v_x and v_y) which means you can solve.

Might be a cleaner way of doing it out there, but thats how I'd do it

Answer 11

you actually don't need the pythagorean equation I included in there... You can ignore the v_i^2 = v_x^2 + v_y^2. It will only over complicate solving

Answer 12

I know the initial velocity of the child would be 71.84 ft/sec. Using a trajectory calculator that I found online to save time, I determined that an 80.9º angle would be required for a 50 foot range at 71.84 ft/sec. However, the resulting peak height equals 78.3 feet.

I now know that my instructor's answer key is most likely incorrect, and even if he had provided us with the proper equations to complete this problem, a launch angle would have made this so much easier to solve.

I'm still going to play around with the equations you provided to see if I can figure this out. I really appreciate your help! Thank you!

Answer 13

good luck and hopefully that works out!