What is her maximum speed? Answer in units ofm/s. Part2: please show work Given:

Question

What is her maximum speed? Answer in units ofm/s. Part2: please show work Given: In thefollowing choices, htop is the
height of the highest point and hbot is the
height of the lowest point.
At what height above the ground will the
girl be moving at a speed half of her maximum
speed?
1. h1/2 =3/4htop
2. h1/2 =1/4(3 htop + hbot)
3. h1/2 =(1/(2^1/2)) htop
4. h1/2 =(1/(2^1/2))[htop + ((2^1/2) 1) hbot]
5. h1/2 =1/4htop
6. h1/2 =1/2(htop + hbot)
7. h1/2 =1/2(htop
8. h1/2 = hbot
9. h1/2 =1/4(htop + hbot)
10. h1/2 = htop Part2: please show work Given: In thefollowing choices, htop is the
height of the highest point and hbot is the
height of the lowest point.
At what height above the ground will the
girl be moving at a speed half of her maximum
speed?
1. h1/2 =3/4htop
2. h1/2 =1/4(3 htop + hbot)
3. h1/2 =(1/(2^1/2)) htop
4. h1/2 =(1/(2^1/2))[htop + ((2^1/2) 1) hbot]
5. h1/2 =1/4htop
6. h1/2 =1/2(htop + hbot)
7. h1/2 =1/2(htop
8. h1/2 = hbot
9. h1/2 =1/4(htop + hbot)
10. h1/2 = htop

Explanation / Answer

Keeping the ground as a reference point. We need to equate initialenergy to final energy. At the highest point, KE=0, PE=mghtop=2.1mg At the lowest point KE=1/2mv2 ,PE=mghbot=0.7mg Equating energies: 2.1mg =1/2mv2 +0.7mg                            1.4mg=1/2mv2                                      v=5.08m/s for velocity to be 5.08/2 =2.54m/s we need to substitute this value in the energy equation tocalculate height. so: 2.1mg=1/2m*(2.54)2+mgh solving..... h=1.77m so, putting the values in, 1/4(3 htop + hbot) turns out to becorrect.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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