Question

A projectile is launched from a cannon; it has an initial speed of 85 m/s at an angle of 33 degrees above horizontal. The muzzle of the cannon is 2.5 meters above the ground. How far down range will the projectile land?

Answer 1

in vertical direction:

vi = 85*sin 33 = 46.3 m/s

a= -9.8 m/s^2

d = -2.5 m

use:

d = vi*t + 0.5*a*t^2

-2.5 = 46.3*t + 0.5*(-9.8)*t^2

4.9*t^2 - 46.3*t - 2.5 = 0

This is quadratic equation (at^2+bt+c=0)

a = 4.9

b = -46.3

c = -2.5

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 2.193*10^3

roots are :

t = 9.50 and t = -5.369*10^-2

since t can't be negative, the possible value of t is

t = 9.50 s

in horizontal direction:

v = 85*cos 33 = 71.3 m/s

t = 9.50 s

So,

range = v*t

= 71.3 m/s * 9.50 s

= 677 m

Answer: 677 m