Suppose you throw a 0.053-kg ball with a speed of 11.2 m/s and at an angle of 33

Question

Suppose you throw a 0.053-kg ball with a speed of 11.2 m/s and at an angle of 33.0° above the horizontal from a building 12.6 m high.

(a) What will be its kinetic energy when it hits the ground?
_________

(b) What will be its speed when it hits the ground?
_________ the answer for A i get 19.28.. but its wrong.. thank you for your help
Suppose you throw a 0.053-kg ball with a speed of 11.2 m/s and at an angle of 33.0° above the horizontal from a building 12.6 m high.

(a) What will be its kinetic energy when it hits the ground?
_________

(b) What will be its speed when it hits the ground?
_________ the answer for A i get 19.28.. but its wrong.. thank you for your help
(a) What will be its kinetic energy when it hits the ground?
_________

(b) What will be its speed when it hits the ground?
_________ the answer for A i get 19.28.. but its wrong.. thank you for your help

Explanation / Answer

The initial speed of the ball u = 11.2 m/s
The mass of the ball is m = 0.053 kg
The angle of projection theta = 33.0 degree
The height of the building h = 12.6 m a) According to the work energy theorem, the kinetic energy of the ball must be equal to change in its potential energy. K = DeltaU K = mgh K = 0.053 kg * 9.8 m.s^2 * 12.6 m K = 6.544 J b) The horizontal component of the ball's velocity vx = ucos33                                                                          = 11.2 cos33                                                                          = 9.393 m/s The vertical component of the ball's velocity is given by vy = sqrt[2gh-(usin33)^2] vy = 14.48 m/s The velocity of the ball before hitting the ground is v = sqrt(vx^2+vy^2)                                                                               = 17.26 m/s

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