Question

A ball with an initial velocity of 5.55 m/s is shot at an angle of 30...

A ball with an initial velocity of 5.55 m/s is shot at an angle of 30 degree above the horizontal using a launcher. The launcher is 1.11m high.

i. How long is the ball in the air, before it hits the ground?

ii. How far from the launcher does the ball land on the ground?

Homework Answers

Answer #1

i)

consider the motion along the Y-direction

Yo = initial position = 1.11 m

Voy = initial velocity = 5.55 Sin30 = 2.78 m/s

ay = acceleration = - 9.8

Yf = final position = 0

t = time of travel in air = ?

using the equation

Yf = Yo + Voy t + (0.5) ay t2

0 = 1.11 + (2.78) t + (0.5) (- 9.8) t2

t = 0.84 sec

ii)

consider the motion along the X-direction

Vox = initial velocity = 5.55 Cos30 = 4.8 m/s

X = horizontal displacement = ?

t = time of travel in air = 0.84 sec

ax = acceleration along horizontal direction = 0 m/s2

using the equation

X = Vox t + (0.5) ax t2

X = (4.8) (0.84) + (0.5) (0) (0.84)2

X = 4.032 m

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