Asked by Kratos
An object is thrown with projectile motion at an initial velocity Vo= 30.4m/s at an angle 30°. Determine the maximum height of the projectile (in meters)
Answers
Answered by
oobleck
at the top, v = 30.4 cos30° = 30.4 * √3/2, and so KE/m = 1/2 v^2 = 30.4^2 * 3/8
PE gained = mgh
initial KE/m = 1/2 * 30.4^2
∆KE/m = 30.4^2 * 1/8 = gh
h = 30.4^2/(8g) = 11.77 m
PE gained = mgh
initial KE/m = 1/2 * 30.4^2
∆KE/m = 30.4^2 * 1/8 = gh
h = 30.4^2/(8g) = 11.77 m
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