Mechanics problem

[Fraggr]

Hey guys, I'm going through my textbook and have no idea how to do this question:

A stone is thrown down the slope as shown. Determine the magnitude and direction (theta) of its initial velocity so that the stone will rise 12m and still have a range of 50m down the slope.

------12m up here

_____ <- stone thrown from left hand side
\......|
.\.....|
..\....|
...\...|
....\..|
.....\.|
......\|

Hypotenuse 50m, top 40m, right hand side 30m. Sorry for the poor diagram.

This means it has to travel 40m in the x direction, and 54m in the y direction (12m up, 12m down, then another 30m down).

The answers are 17.74m/s at 96.7 degrees.

If I'm given one of those answers the question is easy, but finding both unknowns is a bit of a challenge.

 

[touch]

96.7 degrees.

Surely that can't be right? Where is 0 degrees? horizontal?

So you're basically throwing a stone at 6.7 degrees off of vertical, it goes up 12m, back down 42m and yet manages to travel horizontally 40m?
Doesnt sound right to me.

 

[wonko]

If the projectile goes 40m horizontally and falls 30m vertically before hitting the slope (so has a range of 50 along the slope),

horizontally it travels 40m so using s=ut (no horizonal acceleration),
40 = (u cos theta) t

vertically using s=ut+0.5at^2
-30 = u sin theta - 0.5 g t^2

If we eliminate t we get an equation with only u and theta in it.

If initial velocity is u, to reach a peak height of 12m, using v^2=u^2+2as
0 = (u sin theta)^2 - 2xgx12

That gives us a value for u sin theta which you should be able to use to solve u and theta... I think.

Edit - not done the sum but if you eliminate u, you get a quadratic in tan theta. Also realised I got the 30m and 40m the wrong way round

 

[Fuzz]

Oh... THAT sort of mechanics problem!

<closes door quietly>