Question

A 6.80-g steel ball is fired from a spring-loaded cannon. The spring (k = 50 N/m) is compressed 9.0 cm before launching the ball at a 45˚ angle. What is the velocity of the ball when launched? How far will it go? What is its kinetic energy at the top of its trajectory?

Answer 1

the kinetic energy of the ball is achieved from spring energy

0.5mv^2 = 0.5kx^2

0.0068*v^2 = 50*0.09^2

v = 7.72 m/s

velocity of ball when launched = 7.72 m/s

Range = U^2 sin 2 / g

       = 7.72^2 * sin 2*45 / 9.8

      = 6.1 m

at the top , the ball will have only x-component of Velocity

            = 7.72 cos 45 = 5.46 m/s

KE at top = 0.5mv^2 = 0.5*0.0068*5.46^2 = 0.1 J