# Launching a ball on a ramp

I am having trouble figuring out this problem:

A ball is launched with a velocity of 10 m/s at an angle of 50 deg to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. a) Does the ball land on the ramp or the plateau? When it lands, what are b) the magnitude and c) the angle of its displacement from the launch point?

The diagram that I was given looked like this

--------/|--------
-------/ |
------/--d2
-----/---|
___o/-d1--

note: please ignore the ---- along the left side of the ramp, it was the only way i could figure out how to create the picture. the o = the ball. it's on the ground. the --- lines to the right of the ramp is the plateau.

I was using the trajectory equation:

y = (tan@)x - gx^2/2(Vocos@)^2
= (tan50)(6) - 9.8(6)^2/2(10cos50)^2
= 2.88 m

Then I found the length of the ramp:

ramp = d = (3.6^2 + 6^2)^(1/2) = 36.5 m

With this in mind, I was thinking that the ball would land on the ramp because 2.88 m does not even come close to clearing the length of the ramp, 36.5 m. So, would the magnitude of the displacement be 2.88 m? Concerning the angle, why would it be any different from 50 deg? I think that I'm getting confused about what they're looking for. Could someone help me understand this, please? Thanks so much

Last edited by a moderator:

Staff Emeritus
Gold Member
First off, your 2.88m is the height of the ball at a horizontal distance of 6m from the launch point.

Second, the length of the ramp would be
6.997m not 36.5 ( you used 36 instead of 3.6 when you calculated your formula)

Besides that, you don't need the length of the ramp.

Now since 2.88m is less than the 3.6m height of the ramp, the ball will land on the ramp.

The displacement will be the distance the ball has traveled horizontally when it hits the ramp. (it will intersect the ramp at some point short of 6 meters horizontal displacement.)

If you are looking the angle from horizontal the ball is traveling when it strikes the ramp, it won't be 50°. It would only be 50° if the ball landed at the same height as it was launched. Since it lands at a higher height, the angle will be different.

My advice would be to determine the equation that describes the ramp then equate this to the trajectory formula to find x where the values for y are the same for each equation.

From this you can get both the horizontal and vertical displacement for the ball when it strikes the ramp and then the angle of the path of the ball at that point.