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If U=53m/s and find the launch angle such that the maximum height of the projectile is equal to three times the horizontal range​

Sagot :

Answer:

t = V0 sin theta / g      time to get max height

h = Voy t - 1/2 g t^2 = Vo^2 sin^2 theta / 2 g          max height

R = V0^2 sin 2 theta / g      range formula

h = 3 R        given

Vo^2 sin^2 theta / 2 g = 3 * V0^2 sin 2 theta / g

sin^2 theta = 6 * sin 2 theta = 12 * sin theta cos theta

tan theta = 12

theta = 85.2 deg

Can check:  

Vx = 8.37

Vy = 99.6     assuming V0 =100

t = 99.6 / 9.8 = 10.2 sec     to reach max height

Sx (total range) = 2 * 10.2 * 8.37 = 170 m

Sy = 1/2 * 9.8 * (10.2)^2 = 510 m

Height = 3 * range

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