Homework Help: Projectile Motion Equation Problem

1. Feb 13, 2015

CaptainSensible

Hi Everyone,

I hope someone can help me, I am attempting this projectile problem, but I keep coming unstuck. Could anyone point my in the right direction?

I have attempted to break down the initial velocity into its component forces etc, but as I do not know the initial velocity, final velocity or time, I'm struggling to find a solution.

Also, this question should only require the Kinematic (SUVAT) equations to work it out.

The answer I got is an initial velocity of 13.14, and a total time of 0.86 seconds? But I am not confident in my answer. Thanks for your help everyone, it's been driving me mad!!

James

2. Feb 13, 2015

andrevdh

How did you get to these answers James?

3. Feb 13, 2015

CaptainSensible

My bad, below are my workings:

First step to find the height at point A: 10*Tan(20) = 3.639

I attempted to consider the vertical component only to derive some more information:

I used the only kinematic equation that isn't time dependent: (V2=U2+2as)

I took a= -9.8 and the displacement to be 3.639 (My first problem is that the displacement should be negative? But this results in trying to square root a negative value).

This gave me the y-component of the initial velocity to be (8.446ms-1), and everything is derived from there.

I am not confident that any of this is correct, and my gut feeling is telling me that I need to find a term in the y-component that can be substituted into the x-component, but I can't seem to find a common value.

**Edit: Sorry, I should also have mentioned that I assumed V to be 0**

Thanks,
James

Last edited: Feb 13, 2015
4. Feb 13, 2015

andrevdh

Problem is the projectile do not arrive at ground level with a zero velocity!

5. Feb 13, 2015

CaptainSensible

This is what I thought, without U, V or t is this problem possible? (Without using some sort of differential equation... ?)

6. Feb 13, 2015

andrevdh

Let's try this one
s = ut +1/2 at2
where u = uy is the y-component of the launched velocity
and for the time we try
ux = 10/t .... do you know where this comes from?
See if you can solve for u that way.

7. Feb 13, 2015

azizlwl

You have to resolve the motion into horizontal and vertical.
Horizontally no force acting on it, thus it retains initial velocity, Newton'law.
Vertically there is force acting on it.
Time and distance identical for both motion,

Last edited: Feb 13, 2015
8. Feb 13, 2015

CaptainSensible

Thanks both,

andrevdh, I can see that you can find t in terms of ux as follows:

S=(u+v/2)t
10=(2*ux/2)t
10=uxt
and rearranged to give you: ux=10/t

I've given it another go, and come up with the following:

Using: S=ut+0.5at2
I've substituted the values back in, and got:
-3.64 = uSin40 * (10/uCos40) - 0.5 * -9.8 * (10/uCos40)2
Solving this has given me: 8.38ms-1

Using the horizontal component, the time would then be:

10=(2*8.38/2) * t
t = 10/8.38
t= 1.2 seconds

I'm certainly more confident with the technique, even if I've not overly confident with the final value...
Thanks for all your help everyone, at least now I can talk to my lecturer with a decent attempt.

James

9. Feb 14, 2015

dean barry

Were you given a definite value for g ?

10. Feb 16, 2015

andrevdh

I think you have swopped the velocity components around? Shouldn't the vertical be cosine and the horizontal sine?
It seems to me the 40o angle of the velocity is with respect to the vertical?

11. Feb 16, 2015

dean barry

You may have to juggle both the value for g and the launch velocity to get the answer you have.
Tricky.