A ball launched from ground level lands 2.8 s later on a level field 42 m away from the launch point. Find the

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A ball launched from ground level lands 2.8 s later on a level field 42 m away from the launch point. Find the

[From: ] [author: ] [Date: 13-09-28] [Hit: ]

Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.direction= ° (above the horizontal)-Vy = g*t/2 = 13.Vx = 42/2.Vi = √[Vy²+Vx²] = 20.Θ = arctan[Vy/Vx] = 42.......

A ball launched from ground level lands 2.8 s later on a level field 42 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.)

magnitude= m/s

direction= ° (above the horizontal)

-

Vy = g*t/2 = 13.72 m/s
Vx = 42/2.8 = 15 m/s

Vi = √[Vy²+Vx²] = 20.33 m/s
Θ = arctan[Vy/Vx] = 42.45°

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