Question

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?

Answer 1

i)

consider the motion along the Y-direction

Y_o = initial position = 1.11 m

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

a_y = acceleration = - 9.8

Y_f = final position = 0

t = time of travel in air = ?

using the equation

Y_f = Y_o + V_oy t + (0.5) a_y t²

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

t = 0.84 sec

ii)

consider the motion along the X-direction

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

X = horizontal displacement = ?

t = time of travel in air = 0.84 sec

a_x = acceleration along horizontal direction = 0 m/s²

using the equation

X = V_ox t + (0.5) a_x t²

X = (4.8) (0.84) + (0.5) (0) (0.84)²

X = 4.032 m