A passenger on a moving train tosses a coin with an initial velocity of (-2.25 m

Question

A passenger on a moving train tosses a coin with an initial velocity of (-2.25 m/s) + 4.04 m/s) (from the point of view of the passenger). The train's velocity relative to the ground is (13.1 m/s). (a) What is the minimum speed of the coin relative to the ground during its flight? (b) At what part in the coin's flight does this minimum speed occur? please explain. (c) Find the initial speed and direction of the coin as seen by an observer on the ground. (d) Use the expression for ymax derived in Example 4-7 to calculate the maximum height of the coin, as seen by an observer on the ground. (e) Repeat part (c) from the point of view of the passenger.

Explanation / Answer

a)V_min=10.85m/s b)t=4.04/9.8=0.412s when it's vertical velocity become zero. c)V_x=10.85m/s V_y=4.04m/s V=10.85 i +4.04 j |V|=sqrt(4.04^2+10.85^2)=11.57m/s direction will be in direction =(10.85 i +4.04 j)/11.57 d)Hmax=4.04^2/(2*9.8)=0.8327m e)V_x=-2.25m/s V_y=4.04m/s V=-2.25i +4.04 j |V|=sqrt(4.04^2+2.25^2)=4.624m/s direction will be in direction =(-2.25 i +4.04 j)/4.624 unit vector

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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