SOMEONE ASKED ๐
A good quarterback can throw a football at 27m/s;m/s (about 60mph;mph ).?
HERE THE ANSWERS ๐
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Maximum range is achieved if the ball is thrown at 45 degrees to horizontal.
Initial vertical V component = (sin 45 x 27) = 19.1m/sec.
Time to max. height = (v/g) = 19.1/9.8 = 1.949 secs.
b) Double it, = 3.898 secs. time in air.
Horizontal launch V component = (cos 45 x 27) = 19.1m/sec.
a) Max. range = (19.1 x 3.898s) = 74.45 metres.
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A) R = Vยฒ/g = 74.4 m = 81.3 yd
B) t = โ2*V/g = 3.9 sec
Source(s): Kinematics of constant acceleration
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