# Simple 2-d projective motion

1. Sep 29, 2010

### cp-svalbard

1. The problem statement, all variables and given/known data

Vi=25.0 $$m/s$$
$$\Theta$$=40.0 above the x-axis

You throw a ball at 40 degrees above the horizon at an initial velocity of 25 m/s

The ball lands at the same height as which it was launched.

Is it possible to find the distance traveled (either vertical or horizontal) or the time it took for the ball to land.

2. Relevant equations

Kinematic equations.

3. The attempt at a solution

Seems like this would be pretty simple. but trying to solve for time of distance always is hampered by the initial and final velocities being the same.

So would there be more information needed to solve this?

2. Sep 29, 2010

### jhae2.718

You need to use the standard projectile motion equations describing the x and y components of position as a function of time.

$$x(t)=v_0\cos(\theta)t+x_0$$

$$y(t)=y_0+v_0\sin(\theta)t-\frac{gt^2}{2}$$

What are your initial conditions? What are your final conditions? How can you use these equations to find the final time, and then the distance traveled?

3. Sep 29, 2010

### cp-svalbard

Yeah but since it lands as the same horizontal as its launch the velocities are the same.

Trying to rearrange the equations creates a situation where

Since Vi = (25 cos (40)i, 25 sin (40)j) m/s
Vf = ((25 cos (40)i, -25 sin (40)j) m/s

Anyway i rearrange the KEs to solve to t or r doesn't make sense.

4. Sep 29, 2010

### jhae2.718

Think in terms of positions, not velocities. What condition must be true if the object reaches the ground?

Last edited: Sep 29, 2010
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