A ball launched from ground level lands 2.7 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

Question

A. Find the magnitude m/s

B. Find the direction ° (above the horizontal)

Answer 1

For a projectile fired at an angle of µº to the horizontal with a velocity of Vm/s, the maximum height reached derives from h = V^2(sin^2(µ))/29 and the distance traveled parallel to the ground derives from d = V^2(sin(2µ))/g

1--Vcos(µ) = 50/2.75 = 18.181818
2--Vsin(µ) = 9.8(2.75) = 26.95
3--Vsin(µ)/Vcos(µ) = 26.95/18.18 = 1.48239
4--(µ) = arctan(1.48239) = 56º
5--Therefore, the initial launch velocity is V = 18.1818/cos(56) = 32.51m/s in the direction of 56º above the horizontal.

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