A 50 kg cannon ball is shot horizontally from a cannon at a speed of 80 ms At this speed its kinetic energy will be joules?

Answer 1

Well it is a question of Projetile Motion. Whenever you project an object at some angle with horizontal it traces a curved path. Suppose 'x' be the angle with horizontal, 'm' be its mass, 'u' be its initial velocity. You might be knowing that any vector can be resolved into horizonal and vertical component. Here horizontal component of velocity will be ucosx and vertical will be usinx. At highest point the vertical component of the velocity will become zero. Hence it will have no velocity in vertically upward direction. It will only have velocity in horizontal direction which will be 'ucosx'.(The horizontal component of initial velocity will not affected by force of gravity and hence it will be constant throughout the motion.) So kinetic energy at highest point is: KE=0.5*m*(ucosx)^2 KE=0.5*50*(200*cos60)^2=25*10000=250000 J

Answer 2

The kinetic energy of the cannon ball at 80 m/s velocity can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass and v is the velocity. Plugging in the values, KE = 0.5 * 50 kg * (80 m/s)^2 = 160,000 Joules. Therefore, the kinetic energy of the cannon ball will be 160,000 Joules.

Answer 3

KE = 1/2 m V2 = 1/2 (50) (80)2 = 25 x 6400 = 160,000 joules.

The direction of the shot doesn't matter.

Answer 4

Kinetic Energy = 1/2 M V2 = (25) (80)2 = 160,000joules.