Question

ball of 600grams is kicked at an angle of 35° with the ground with an initial velocity
a) what is the initial velocity of the ball if the kinetic energy is 22joules when its height is maximum? b) what is the maximum height reached by the ball​

Answer 1

Answer:

a) 10.45 m/s

b) 1.8 m

Explanation:

a) Kinetic Energy at maximum height = 1/2 x mass x (velocity at maximum height)^2

At h max, the velocity would have a y component of 0,

Kinetic energy at h max = 0.3 Kg x (v_x)^2 where v_x is the x component of the velocity

22 J = 0.3 Kg x (v_x)^2

(v_x)^2 = 73.33 = v_x = 8.56 m/s

v_x at the initial position = 8.56 m/s because v_x remains constant

Resultant is at 35 degrees

tan(35) = v_y sin theta/ v_x + v_y cos theta

theta = 90 becuase v_x and v_y ate perpendicular

tan(35) = v_y/v_x

tan(35) = v_y/8.56

v_y = tan(35) x 8.56 m/s

v_y = 0.7 x 8.56 m/s = 6 m/s

Resultant velocity of v_x and v_y = initial velocity = sqrt(v_x squared + v_y squared)

Resultant velocity = 10.45 m/s

b) Conserving energy:

1/2 m (v_y)^2 = mgh

1/2 (v_y)^2 = gh

h = (v_y)^2/2g

v_y = 6 m/s

h = 6^2/20

h = 36/20

h = 1.8 m