Projectile Motion

**vesperka** · November 26th 2007, 09:48 AM

We have a cannon that shoots out a ball, and we have to make it hit a target on the floor.

We have to aim our cannon 25.0 degrees, and it is placed on a ledge 1.51m off the floor. The ball has an initial velocity of 4.534m/s. I think we have to find the x and y component vectors, but I already tried solving it and launching it and we overshot it (our guess was 1.62meters).

**vesperka** · November 26th 2007, 10:11 AM

For me, first I resolved the two component vectors. The hypotenuse is 4.534, so using SOH CAH TOA I found the Y component to be 1.916m/s and the X to be 4.109m/s. Then, I used the distance formula

Dy = VoT + 1/2(g)(t^2)

and solved for T to get

1.51 = (1.916)(t) + 1/2(9.80)(x^2)

Then T = .393

Then to solve for Dx I used

Dx = VoT

the acceleration part of the equation is eliminated because their is constant uniform velocity on the X component vector.

Dx = (4.109)(.393)
Dx = 1.62m

This answer is wrong though

**topsquark** · November 26th 2007, 10:52 AM

Originally Posted by vesperka

We have a cannon that shoots out a ball, and we have to make it hit a target on the floor.

We have to aim our cannon 25.0 degrees, and it is placed on a ledge 1.51m off the floor. The ball has an initial velocity of 4.534m/s. I think we have to find the x and y component vectors, but I already tried solving it and launching it and we overshot it (our guess was 1.62meters).

The reason it isn't working for you is you are setting it up wrong.

I have a coordinate system with an origin at the point directly below (1.51 m below) where the cannon ball is fired. I have a +x direction to the right (the direction I'm firing the ball in my diagram) and a +y direction upward.

Here's what we know:
$t =$ ? <-- This is the time when the ball hits the floor.
$x_0 = 0~m~~~~x =$ ? $y_0 = 0~m~~~~y = 0~m$
$v_{0x} = v_0~cos(25.0)~m/s~~~~v_x = v_{0x}~~v_{0y} = v_0~sin(25.0)~~~~v_y =$ ?
$a_x = 0~m/s^2~~~~a_y = -9.8~m/s^2$

We want x, so
$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

$x = v_{0x}t$
which we can't solve for since we don't know t.

So
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$0 = 1.51 + v_{0y}t - 4.9t^2$

We can solve this for t:
$t = \frac{-1.91615 \pm \sqrt{1.91615^2 - 4 \cdot -4.9 \cdot 1.51}}{2 \cdot -4.9}$

$t = 0.784078~s$
disregarding the negative solution.

So we have:
$x = v_{0x}t = 3.22193~m$

Always organize your data before you start working the equations!

-Dan

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