Projectile Launching Problem. Solve with kinematics

[DeerHunter]

Homework Statement

A cow, with a mass of 327 kg, if fired from a medieval catapault, and travels a horizontal distance of 1375 m. It lands in a depression 39 m below its starting position. If it is launched at an angle of 37.0\circ above the horizontal, find its initial speed.

Givens
HOR
\Delta dx= 1375
V1x= ?
ax= 0m/s²
v2x=v1x

VER
\Delta dy= -39 m
v1y= ?
ay= -9.81m/s²
v2y=?

\Delta T=?

Homework Equations

<br /> v^2 = v_0^2 + 2 a \Delta x<br />
<br /> x = x_0 + v_0 t + (1/2) a t^2<br />
<br /> v^2 = v_0^2 + 2 a \Delta x<br />

The Attempt at a Solution

I multiplied my vertical and horizontal displacements by 39.0* (i.e. dx= 1375*cos37) but I am not sure if this is right or if I should only use that for the velocity's only. Also I am not sure where to go from here all relevant equations require at least v1x,v2x,v1y,v2y and/or t. One last thing, I am not sure how mass is to be used as I am suppose to solve with kinematics, not forces but I could be wrong. Any help is appreciated! Thanks!

 

[quanticism]

"I multiplied my vertical and horizontal displacements by 39.0* (i.e. dx= 1375*cos37)"

I don't really understand what you're doing here. If you denote the initial speed as V and you know the angle of launch is 37 degrees, then the initial horizontal component of velocity is Vcos37. Likewise, the initial vertical component is Vsin37. I don't see why you need to multiply the range(horizontal displacement) by cos37.

Oh, and in your "Relevant equations", I think you meant your first eqn to be: v = v_0 + at.

As with most projectile motions questions, you use those equations on the x and y directions to obtain 2 expressions. You can then link these expressions together to obtain your answer.

I think the working out is going to get rather messy for this particular question though.

 

[DeerHunter]

I don't really understand what you're doing here. If you denote the initial speed as V and you know the angle of launch is 37 degrees, then the initial horizontal component of velocity is Vcos37. Likewise, the initial vertical component is Vsin37. I don't see why you need to multiply the range(horizontal displacement) by cos37.

I am sorry, after I posted this I realized that this would do nothing. I just tried something because I was running out of ideas. My only other idea would be to use this equation for y as max height for v2y is 0 m/s. (Correct me if this is wrong)
<br /> <br /> v1y^2 = v_0^2 + 2 a \Delta x<br /> <br />

but eliminate v_0^2.

 

[quanticism]

My hint would be to use:

v^{2} = u^{2} + 2(a)(s)
v = u + at

(Here , v = final velocity, u = initial velocity, a = acceleration, s = displacement)

on the vertical direction to find an expression for the time of flight. Alternately, you could use equation 2 directly to find the time but you'll need to solve a rather messy quadratic.

Then use equation 2 on the horizontal direction and everything should fall into place.

I believe the working out will become messy so if anyone has more elegant ways of doing it, please post.

 

[DeerHunter]

My hint would be to use:

v^{2} = u^{2} + 2(a)(s)
v = u + at

(Here , v = final velocity, u = initial velocity, a = acceleration, s = displacement)

on the vertical direction to find an expression for the time of flight. Alternately, you could use equation 2 directly to find the time but you'll need to solve a rather messy quadratic.

Then use equation 2 on the horizontal direction and everything should fall into place.

Thank you very much!
What I did for here was make a equation for time of flight
dx=v1*cos37* t +1/2 at^2
1375=v1x*cos37*t
\Delta t= 1372/v1*cos37


now that i have \Delta t I have inserted it into my vertical equation so
dy=v1y*cos37* t +1/2 at^2
-39=v1*sin37*(1372/v1*cos37) + 1/2 (-9.81)(1372/v1*cos37)^2
Vsin37= -4.905(1372/v1*cos37)^2(-39) /(1372/v1*cos37)


I'm going to play around with this and update with new info. You're right in saying this is a messy equation haha. Thanks again!