Question

You throw a ball for your dog to fetch. The ball leaves your hand with a speed of 2 m/s, at an angle of 30◦ to the horizontal, and from a height of 1.5 m above the ground. The mass of the ball is 0.5 kg. Neglect air resistance in what follows.
(a) How high does the ball rise above the ground?
(b) What is the kinetic energy of the ball as it hits the ground?
(c) Now let the system be the ball alone. How much work does the Earth do on the ball while it is
in flight? (from start to finish)
(d) What is the velocity of the ball as it hits the ground? Report it as a vector.
(e) How far away from you (horizontally) does the ball land?

Answer 1

Here ,

u = 2 m/s

theta = 30 degree

h = 1.5 m

a) height risen = 1.5 + u^2/(2g)

height risen = 1.5 + 2^2/(2 * 9.8)

height risen = 1.70 m

b)

kinetic energy at ground = 0.50 * 0.50 * 2^2 + 0.50 * 9.8 * 1.5

kinetic energy at ground = 8.35 J

c)

work done by earth = 0.50 * 9.8 * 1.5

work done by earth = 7.35 J

d)

for the speed at ground

0.50 * 0.50 * v^2 = 8.35

v = 5.8 m/s

for the vertical velocity

(2 * cos(30))^2 + vy^2 = 5.8^2

vy = 5.54 m/s

velocity = 2* cos(30) i - j * 5.54 m/s

velocity = 1.73 i - 5.54 j m/s

e)

let the time of flight is t

-1.5 = 2 * sin(30) * t - 0.50 * 9.8 * t^2

solving for t

t = 0.665 s

horizontal distance = 0.665 * 2 * cos(30)

horizontal distance = 1.15 m