Question

A con on at ground level launches a cannonball horizontally. If the cannon fires with a velocity of 28.3m/s and at an angle of 30 degrees how far will the cannonball travel?

Answer 1

v_o = initial velocity = 28.3 m/s

= angle of launch = 30

consider the motion along the vertical direction or Y-direction

V_oy = initial velocity In Y-direction = v_o Sin = (28.3) Sin30

a = acceleration = - 9.8

Y = displacement = 0 m

t = time of travel

using the equation

Y = V_oy t + (0.5) at²

0 = ((28.3) Sin30) t + (0.5) (- 9.8) t²

t = 2.88 sec

consider the motion along the horizontal direction or X-direction

V_ox = initial velocity In X-direction = v_o Cos = (28.3) Cos30

a = acceleration = 0

X = displacement = ?

t = time of travel = 2.88

using the equation

X = V_ox t + (0.5) at²

X = ((28.3) Cos30) (2.88)+ (0.5) (0)(2.88)²

X = 70.6 m

t = 2.88 sec