velocity of a projectile?

Question

show that at any time t, where friction is negligible, the velocity of a projectile (launced at an angle Ө with a speed V1) makes an angle Өt with the horizontal given by the expression

Өt= tan -1=(tan = (gt/v initial cos Ө))
where g<0

really just looking for someone to help walk me through this I have not been able to get ahold of my physics professor the past two days.

Answer 1

When you launch the projectile a a velocity V1 and an angle theta, you can describe its motions in terms of x and y, x being the horizontal and y being the vertical.

The initial horizontal velocity Vx is V1*cos(theta), and the initial vertical velocity is V1*sin(theta). Review your trig if you don't see why.

The is (I assume) no acceleration in the x direction, so the initial horizontal velocity Vx is always the aforementioned value.

In the y-direction, gravity will change the the vertical velocity Vy, according to Vy = V1*sin(theta) + gt, where g = -9.8m/s^2.

Drawing these vectors to get a resultant V vector, we can see that there is an angle Өt. Tan (Өt) = Vy/Vx = (V1(sin(Ө)+gt)/(V1*cos(Ө)). Take the inverse tangent of both sides to get Өt. I am worried because this is correct as I know it, but it does not match up with what you have written up. Are you sure you typed it correctly?

If later today I notice I made a mistake, I will come back and update.

As a word of advice, please, please don't just copy my solution. First, I am not a god who is always right. Second, neither I nor anyone else can take the midterm for you. So do learn it as you do it.

Answer 2

The text, University Physics by Sears,Zemansky, and Young has this worked out, should be in the library.