Aprojectile of mass 2 kg is fired with a speed of 20 m/s at an angle of 30° with respect to the horizontal. (a) calculate the initial total energy of the projectile given that the reference point of zero gravitational potential energy at the launch position. (b) calculate the kinetic energy at the highest vertical position of the projectile. (c) calculate the gravitational potential energy at the highest vertical position. (d) calculate the maximum height that the projectile reaches. compare this result by solving the same problem using your knowledge of projectile motion.

Explanation:

a ) Initial kinetic energy

1/2 mv²

= .5 x 2 x 20 x 20

= 400 J

Total energy = P E + KE

0 + 400 = 400 J

b )

Velocity at the highest point will be equal to horizontal component of initial velocity

= 20 cos 30

= 17.32 m/s

Kinetic energy at the highest point

= ,5 x 2 x 17.32²

= 299.98 J

c )

gravitational potential energy at the highest vertical position

= initial total energy - kinetic energy at the highest point

= 400 - 299.98

= 100 J

d )

PE = mgh

h = PE / mg

= 100 / 2 x 9.8

= 5.1 m

e ) height achieved by a projectile

= u²sin²θ / 2g

= 20 x 20 x sin²30 / 2 x 9.8

= 5.1 m

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.