Projectile Motion Velocity Question

[Procrastinate]

The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is 0.4 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to whioch she rises above the end of the track.

I have tried again and again to do this question but it will not work.

Found initial velocity in the y direction - 4.01.

Then I just substituted these values into the equation: v² = u² + 2as.

I got 0.832.

I subtracted this from 0.4 and got about 0.42m. Wrong, wrong, wrong, that was.

 

[EinsteinKillr]

good news, you correctly identified the initial velocity in the y direction.

don't forget that acceleration in your equation is negative g, because gravity is slowing down the object.

Your maximum height will be the moment that v(final) = 0, so...

v(final)^2 = v(init)^2 + 2as
0^2 = 4.01^2 + 2(-9.8)s

solve for s, and don't forget to add that to your initial height when you leave the ramp, and you will have your answer

 

[Procrastinate]

I got 0.42 (approximately) but that doesn't seem like the right technique. Any other ideas on how to tackle this question?

I have tried using PE and KE to figure this out but I get the same answer.

 

[Epistemic]

Since the question asks you to

find the maximum height H to which she rises above the end of the track.

and states

When she leaves the track, she follows the characteristic path of projectile motion.

you can use either the constant acceleration formulas both you and EinsteinKillr stated, or the energy formula you stated. Both will work in this case. The important thing to realize here, is that they are asking for the height above the end of the track, so you won't need to add or subtract anything from the height you get from either of those techniques.

If these techniques don't seem correct to you, might you state why? Also if you have the given answer, stating what it is might help identify which technique needs to be used.

 

[rwisz]

It seems like you are on the right track, but there are a few things that need to be corrected in your calculations. Firstly, the initial velocity in the y-direction should be found using the sine of the angle, not the cosine. So the correct initial velocity in the y-direction would be 5.4*sin(48°) = 3.94 m/s.

Secondly, the equation you used, v^2 = u^2 + 2as, is for finding the final velocity, not the maximum height. To find the maximum height, we need to use the equation v^2 = u^2 + 2gh, where g is the acceleration due to gravity (9.8 m/s^2).

So, plugging in the values, we get 0 = (3.94)^2 + 2*(-9.8)*h. Solving for h, we get a maximum height of 0.76 m above the end of the track.

I would recommend double checking your calculations and equations to ensure accuracy. Remember to always use the appropriate equations for the given scenario. Good luck!