Question

A cannon fires a cannonball at some angle with respect to the horizontal. It travels a horizontal distance of 26.9 meters and hits the ground in 2.5 seconds. Assume that it lands at the same level as at which it was fired and that there is no air drag.

what is the initial speed at which the cannonball was fired?

Answer 1

Range of projectile

R = V^2*sin 2A/g

time of flight

T = 2V*sin A/g

Given R = 26.9 m

T = 2.5 sec.

using these values

26.9*g = V^2*sin 2A = V^2*2*sin A*cos A

(V*cos A)*(V*sin A) = 26.9*g/2

by equation 2

2.5*g/2 = V*sin A

V*sin A = 1.25*9.81 = 12.2625

V*cos A = 13.45*g/1.25*g = 10.76

Now we have

V*sin A = 12.2625

V*cos A = 10.76

adding both of them after square

V^2*sin^2 A + V^2*cos^2 A = 12.2625^2 + 10.76^2

sin^2 A + cos^2 A = 1

V^2 = 12.2625^2 + 10.76^2

V = sqrt(12.2625^2 + 10.76^2) = 16.31

Velocity = 16.31 m/sec