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Homework Statement
During baseball practice, you go up into the
bleachers to retrieve a ball. You throw the
ball back into the playing field at an angle of
42° above the horizontal, giving it an initial
velocity of 15 m/s. If the ball is 5.3 m above
the level of the playing field when you throw
it, what will be the velocity of the ball when
it hits the ground of the playing field?
<=42°
vi=15m/s
h=5.3m
vf=?
Homework Equations
h=-0.5gt^2+VosinθΔt
vf=vi+aΔt
The Attempt at a Solution
5.3m=-0.5(9.8)t^2+15(sin 42°)Δt
t=1.02s
vf=(15)(sin 42°)+(-9.8)(1.02)
vf=0.041m/s
vx^2 + vy^2 (square root)
(15 x cos 42)^2 + 0.041^2
=11.15m/s
what am i doing wrong? Answer is 18m/s.
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