A cannon ball is launched (from the ground) at an angle of 35 degrees, and lands 2500 m from where it was launched. At what speed was the cannon ball launched?

1 Answer
Apr 5, 2016

v_i=161,55 m/s

Explanation:

v_i:"initial velocity"
alpha:" angle"
g=9,81 m/s^2
x_m=2500 m

x_m=(v_i^2*sin2 alpha)/g

2500=(v_i^2*sin2*35)/(9,81)

v_i^2=(2500*9,81)/sin 70

v_i^2=24525/(0,9396926208

v_i^2=26098,96

v_i=161,55 m/s