# Projectile Motion - Acceleration and Angles

1. Aug 16, 2012

### jnrn

Finding the total distance a jaguar has jumped from a 30° angle at 3.33m/s.

I've split the jump into horiz./vert. components:
Vertical: v sin(θ) = 3.33sin(30) = 1.67 ms-1
Horizontal: v cos(θ) = 3.33cos(30) = 2.88 ms-1

I am just unsure as to how to find the time taken to reach maximum height using the vertical value in order to sub into the horizontal etc.

Any help is greatly appreciated!

2. Aug 17, 2012

### andrien

the time is 2vsinθ/g,multiply by horizontal velocity to find distance.

3. Aug 17, 2012

### CAF123

Just to be a little clearer, the equation $t = \frac{2vsinθ}{g},$ tells you how long it takes for the projectile to cover the whole range.
However, in answer to your question about finding the time to maximum height, you just divide this expression by 2 since a parabola has symmetry about the apex.

Or it can be derived: $$v_y = v_{oy} - gt$$
At the top of flight, $v_y = 0 => t = \frac{v_oy}{g} = \frac{v_o sinθ}{g}$