u = 27 cos 45 forever
Vi = 27 sin 45 at start
h = Vi t -4.9 t^2 = t (Vi -4.9t)
h = 0 at ground
so at ground at t = 0 of course and at
t = 27 sin 45 /4.9
range = u t
max height at t = half that t
h = Vi t - 4.9 t^2
second question nonsense, from where
A baseball hit just above the ground leaves the bat at 27 m/s at 45∘ above the horizontal.
A)How far away does the ball strike the ground?
B)How much time is the ball in the air?
C)What is its maximum height?
Find the time for the ball to go the horizontal distance to home plate. Then find how far the ball drops in this time.
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