In Figure 11-41, a 0.380 kg ball is shot directly upward at initial speed 41.5 m

Question

In Figure 11-41, a 0.380 kg ball is shot directly upward at initial speed 41.5 m/s. What is its angular momentum about P, 2.10 m horizontally from the launch point, when the ball is at the following heights? (a) halfway back to the ground kgm2/s What is the torque on the ball about point P due to the gravitational force when the ball is at the following heights? (b) its maximum height Nm (c) halfway back to the ground Nm I am looking for numerical values for a) b) and c) and an explanation of the work please. THANK YOU

Explanation / Answer

a) conservation of energy. the maximum height is H mgH=mv^2/2 so H=v^2/2g=87.7(m). so velocity of the ball half way. mgH=mv0^2/2+mgH/2 so v0=29.3(m/s) so the angular momentum. v0*2.1*m=23.4(kgm2/s) ----- b) and c) its always equal to 2.1*m*g. the reason is because the horizontal distance is unchanged and equal to 2.1m. c)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.