Projectile Motion from a Cliff: Solving for Initial Velocity and Launch Angle

[pkc111]

Homework Statement

A projectile fired up into the air from the top of a 75m high cliff hits the ground 500m out from the base.
Calculate the initial launch velocity (U), and launch angle (theta) of the projectile.

Homework Equations

1) v=u+at
2) v²=u² +2as
3) s=ut+0.5at²

The Attempt at a Solution

Consider up to be positive direction.
Consider vertical component (y component) of motion for entire flight:
a=-9.8 m/s²
s=-75m
u_y=Usin(theta)

Sub into Eqn 3:
-75=Usin(theta)t+0.5(-9.8)t² #EQN A

Consider horizontal component (x component) of motion for entire flight:
a=0
s=500m
u_x=ucos(theta)

Sub into Eqn 3:
500=ucos(theta)t #EQN B


Now comes the problem... I have created 2 equations (A,B), but with 3 unknowns (u, theta,t). I think I need another equation ?

This is REALLY bugging me, please help.
Thanks very much.

 

[kuruman]

Your thinking is correct. You do need more information. The answer to this problem is not uniquely defined with the information that is given.

 

[pkc111]

OK thanks, but the book gives the answer as 65m/s at an angle of 40degrees to the horizontal.
Could it be that I need to create another equation from my data ??

Thanks

 

[kuruman]

As I said earlier, there are pairs of angles and initial speeds that give the same numbers. The numbers that are given by the problem do not uniquely define a solution. Try this:

Assume that the angle is 40^o and that the initial speed is 65 m/s.
Solve your EQN B for the time of flight.
Plug the numbers in the right side of EQN A; you should get -75 m.

Now repeat but with an angle of 60^o and initial speed 72.15 m/s and see what you get.

The answer to this problem is not unique unless some other quantity is given to remove the ambiguity. You have created all the equations that can be created from what is given. To get another equation, you need more data.

 

[pkc111]

Thanks kuruman I see your point.