Question

A canon is firing a projectile. Select an angle for a launch. Choose an angle between 15 and 75 degrees.

Initial speed is 22 m/s. Mass is 7.3g. Diameter of projectile is 0.25m.

Determine the vertical and horizontal components of the projectile.

Calculate the range of the projectile. Write down the equation used to solve for the range.

Answer 1

initial speed=v=22 m/s

mass=m=7.3 gm=7.3*10^(-3) kg

let the angle be theta.

then vertical component of initial speed=22*sin(theta)

horizontal component=22*cos(theta)

in horizontal direction,acceleration is 0.

so velocity will remain unchanged.

in vertical direction,acceleration is 9.8 m/s^2,downwards

so velocity at time t is given by :

22*sin(theta)-9.8*t

now time taken to reach ground=2*time taken to reach the maximum height

at maximum height,vertical component is 0.

so t=22*sin(theta)/9.8

hence range=22*cos(theta)*2*22*sin(theta)/9.8

=484*sin(2*theta)/9.8