physics4
posted by Anonymous .
A baseball is thrown horizontally from a height of 11.20 m above the ground with a speed of 27.5 m/s. Where is the ball after 0.90 s have elapsed?
The ball is 1 a horizontal distance of 2 m from the launch point.

Xo = 27.5 m/s.
Yo = 0.
T = 0.9 s. = Time in flight.
Dx = Xo*T = 27.5m/s * 0.9s = 24.75 m.
from launch point.
h = ho _ 0.5g*T^2,
h=11.2  4.9*(0.9)^2 =11.23.97=7.2 m.
above gnd.
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