Asked by Dejen
Find the range of projectile launched at the range of 45°with an initial of 25m/s ?
Answers
Answered by
Anonymous
horizontal speed = u = 25 cos 45 = 17.7 m/s
so
range = 17.7 T where T is time in air
Initial vertical speed= 25 sin 45 = Vi = 17.7
v = Vi - 9.81 t
when v = 0 at top, we are halfway there
t = 17.7 / 9.81 = 1.95 seconds to top
so T = 2 t = 3.9seconds
so
range = 17.7 * 3.9 = 69 meters
so
range = 17.7 T where T is time in air
Initial vertical speed= 25 sin 45 = Vi = 17.7
v = Vi - 9.81 t
when v = 0 at top, we are halfway there
t = 17.7 / 9.81 = 1.95 seconds to top
so T = 2 t = 3.9seconds
so
range = 17.7 * 3.9 = 69 meters
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