PHYS 2141, Fall 2017, Homework #5 1. A ball is thrown upward at an 2. (Coletta,

Question

PHYS 2141, Fall 2017, Homework #5 1. A ball is thrown upward at an 2. (Coletta, Ch 3, Prob 21) A runner moving at a constant speed of 10.0 m/s rounds a curve of radius 5.00 m. Compare 3. (Coletta, Ch 3, Prob 25) Find the speed of a lunar orbiter in a circular orbit that is just above the surface of the moon, angle of 30° to the horizontal and lands on the top edge of a building that is 20 m away. The top edge is 5.0 m above the throwing point. How fast was t the acceleration of the runner. Are these numbers realistic? Ans. 20.0 m/s given that the orbiter's acceleration is equal to the moon's gravitational acceleration of 1.67 m/s he ball thrown? Ans. 20 m/s Ans. 1.70 × 103 m/s

Explanation / Answer

1)

Horizontal Component

The horizontal component of the velocity is constant as it is perpendicular to g.

vcos30=20/t

v*0.866=20/t

t=20/0.866v ................(1)

Vertical Component

s = ut+1/2at^2

5 = vsin30t1/2gt^2

5 = v*0.5*t1/2*9.8*t^2

5 = v*0.5*t4.9t^2 ......................(2)

from equ. 1 and 2 ....

5 = v*0.5*20/0.866v 4.9t^2

5 = 10/0.8664.9t^2

4.9t^2 = 11.5475 = 6.547

4.9t^2 = 6.547

t^2 = 6.547/4.9=1.336

t = 1.336=1.156 s

from equ. 1

20/0.866v = t

0.866v=20/t = 20/1.156

v = 20/(1.156*0.866) = 19.97 m/s

2)

for centripetal acceleration:

a = v^2/r

a = 10^2/5 = 20 m/s^2

3)

for centripetal acceleration:

a = v^2/r

r = 1737 km

v^2 = a*r

v^2 = 1.67*1.737*10^6 m^2/s^2

v^2 = 2.90079*10^6 m^2/s^2

v = 1.703*10^3 m/s

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