# Projectiles launched from an angle

1. Oct 17, 2004

### alanna

I know how to solve projectile problems when they are launched from an angle if I know the angle and the initial velocity, I need to know:

a. Can you use the vertical displacement (m) and the initial velocity(m/s) to find the angle?

b. How do you find the initial velocity when you know the total horizontal displacement and the total Vertical displacement and the angle at which it is launched? (i.e. a ball is launched at a 35.0 degree angle and it clears a 21.0m wall 130.0m away, what is the initial velocity at which the ball was thrown?)

2. Oct 18, 2004

### LENIN

Answer A

1. Yes you can.

2.Try this way

v_end^2=v_h^2+2a(x-x_0)
0=v_h^2+2g(0-h)
(2g(0-h))^1/2=v_h

sinX=V_h/initial velocity

3. Oct 18, 2004

### alanna

But I don't know the Final velocity or the initial velocity of the horizontal.

4. Oct 18, 2004

### Pyrrhus

The Horizontal has constant speed. Vx is constant.

The Velocity of the projectile motion can be done with its components, where component x is an uniform speed movement, and component y will experience constant acceleration movement.

Last edited: Oct 18, 2004
5. Oct 18, 2004

### Pseudo Statistic

For b, the attached picture should be a correct method of finding the initial velocity, using the appropriate formulae of motion. (v^2 - u^2 = 2as and basic trigonometric..)

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6. May 1, 2011

### cassbass2

Thank you so much, I've been stuck on a problem like this forever.

7. May 2, 2011

### George302

sorry but what do u mean when u use v_h?

vertical velocity? vertial height (S_y)?

Sin x = ? U_y / U_x ????

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