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CollegeStudent
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Homework Statement
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?
Homework Equations
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?