In the figure, a 0. 400 kg ball is shot directly upward at initial speed 22. 8 m

Question

In the figure, a 0. 400 kg ball is shot directly upward at initial speed 22. 8 m/s. What is the magnitude of its angular momentum about P, 1. 07 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the magnitude of the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground? Number Units kg. m^2/s Number Units kg. m^2/s Number Units N. m Number Units N. m

Explanation / Answer

a. at amximum height, velocity is zer0 so, ang momentum = 0 b. half way, velocity =- 27.92 m/s j L= m(rXv ) = 0.4 X ( 1.07 i X -27.92 j) = -11.94 k c. torque = FXR = mg X 1.07 = 0.2744 d. torque remains same as the force nd distance r emains the same

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