# Projectile Motion problem

1. Mar 19, 2013

### CollegeStudent

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

A ball is thrown at a speed of 22.2 m/s at an angle of 44.4° on to a flat level field. The ball leaves the thrower’s hand at a height of 1.88 m. (a) How long is the ball in the air, (b) how far from where the thrower stands does it hit the ground, and (c) what is the magnitude and direction of the balls velocity the instant before it strikes the ground?

2. Relevant equations

3. The attempt at a solution

well first thing i did was break the projectile into it's horizontal and vertical components

22.2cos44.4 = 15.86129349
22.2sin44.4 = 15.53252616

so

(a) How long is the ball in the air

I used

Vf² = Vi² + 2a * Δy
we want Δy so

Δy = (Vf² - Vi²) / 2a

we know the ball starts 1.88m up so

y = ((Vf² - Vi²) / 2a) + 1.88

y = ((0 - 15.53252616²)/2(-9.8)) + 1.88

this came to 2.672475824

this is the max height it reaches

after this...I'm kind of stuck...I believe now i would use

Δy = Vi * t + 1/2 a * t²

to solve for t?

2. Mar 19, 2013

### voko

Yes, that would be correct. What is Vi, assuming the ball is at the max height?

Note you could have used that formula from the beginning. You know that the ball has to drop MINUS its original height to strike the ground, so MINUS height is the LHS side of the equation.