Projectile Motion Problem Involving Initial Velocity and Gravity Only

[Ascendant0]

Homework Statement

A man slides along a water slide that launches him into the air to land into a water pool. The flight distance D = 20m, flight time t = 2.5s, and launch angle is a = 40 (degrees). Find the magnitude of the velocity at launch and landing

Relevant Equations

na

Ok, so to start, the easy part, the velocity in the horizontal x direction (vx):

vx = 20m/(2.5s) = 8m/s

So far so good, but then for the vertical y velocity (vy), my answer is different than the book. I use the equation for position in the y direction, set the y value to 0 (the 2.5s it takes for him to end up back at height y = 0 as he hits the water pool):

y = (vy)t - 1/2gt^2 ---> 0 = vy(2.5s) - 1/2(9.8)(2.5)^2 ---> vy = 12.25m/s

So, with the initial velocity vx and vy, I then calculate the total velocity v by:

v = sqrt(vx^2 + vy^2) ---> v = sqrt(8^2 + 12.25^2) = 14.6m/s

However, their velocity at launch is showing 10.4m/s, but I can't see what I'm doing wrong with my math?

 

[Ibix]

Is there any information that you haven't used? And is there an assumption you've made to avoid using it?

 

[Ascendant0]

Is there any information that you haven't used? And is there an assumption you've made to avoid using it?

The angle of 40 degrees I haven't used, but for how I calculated the x and y axis separately based on times and locations, I don't see how it is relevant to either of them, so if you're referring to the angle, I'm not seeing where it is required for how I calculated it above?

With the x axis, I know he travels the 20m in 2.5s, and the answer in the book has the 8m/s I put in there.

With the y axis, I know he is in air for 2.5s before he gets back to y = 0, and the only acceleration is -g, so again with this one, I don't see how the angle would be relevant to me determining initial velocity based on how long he's in the air before he returns to his original height y = 0?

I appreciate the tip, but I'm still not seeing what I'm doing wrong here?

 

[haruspex]

The angle of 40 degrees I haven't used, but for how I calculated the x and y axis separately based on times and locations, I don't see how it is relevant to either of them, so if you're referring to the angle, I'm not seeing where it is required for how I calculated it above?

But you did assume launch and landing are at the same height. So don’t assume that and use the angle.

 

[Lnewqban]

.. I appreciate the tip, but I'm still not seeing what I'm doing wrong here?

Try the Pythagorean theorem as verification of your calculations, using the launch angle this time.

y = (vy)t - 1/2gt^2 ---> 0 = vy(2.5s) - 1/2(9.8)(2.5)^2 ---> vy = 12.25m/s

Also, isn't the physical meaning of this equation that a projectile launched vertically at 12.25 m/s takes 2.5 seconds to reach the maximum height, point at which Vy is zero?

 

[haruspex]

isn't the physical meaning of this equation that a projectile launched vertically at 12.25 m/s takes 2.5 seconds to reach the maximum height, point at which Vy is zero?

No, that equation says what the OP intends, that the start and finish heights are the same. The equation you are thinking of is ##v_y=gt##.

 

[Ascendant0]

Thanks guys, now I see what I was doing wrong. The heights are different for where he launches from and where he lands. I got it now, I appreciate the help!

 

[brotherbobby]

I'd like to ask you @Ascendant0 to learn ##\rm{\LaTeX}## in order to type your equations better. There is a page on here that will help you.
I'd also like you to show us your final solution to the problem so that, for the moment, we can close the subject.

 

[Ascendant0]

I'd like to ask you @Ascendant0 to learn ##\rm{\LaTeX}## in order to type your equations better. There is a page on here that will help you.
I'd also like you to show us your final solution to the problem so that, for the moment, we can close the subject.

Yes, I do know I need to learn LaTeX. I did glance at it the other day, but I haven't had the time to delve into it yet, and this weekend I have the kids, so not sure when I can, but I am absolutely planning to learn it.

As far as the final solution, it was actually a lot easier than I first thought. Simply take the initial velocity in x and the angle of 40 degrees to get:

vy = 8*tan(40) = 6.7m/s

Then, the 6.7m/s in vy, and 8m/s in vx make a magnitude of v = 10.4m/s at the start

For the finish, vy is -13.9m/s, and vx is still 8m/s, so final v = 16m/s

Again, I greatly appreciate everyone's help with this

 

[brotherbobby]

In your problem statement, you wrote "The flight distance D = 20m".

What distance is this? The horizontal distance or vertical? Or is it the distance along the path?

 

[Ascendant0]

In your problem statement, you wrote "The flight distance D = 20m".

What distance is this? The horizontal distance or vertical? Or is it the distance along the path?

Horizontal distance. My error was in assuming the landing height and the launch height were equal, which they are not. Once that was pointed out to me, I saw exactly where I was going wrong and got it no problem from there. I just need to pay closer attention to exactly what is given (and NOT given) in a problem in the future

 

[brotherbobby]

Horizontal distance. My error was in assuming the landing height and the launch height were equal, which they are not. Once that was pointed out to me, I saw exactly where I was going wrong and got it no problem from there. I just need to pay closer attention to exactly what is given (and NOT given) in a problem in the future

Ok so the horizontal distance from the point of projection to the point of landing is 20m?