Sam heaves a bowling ball with weight 16-lb straight upward, giving it a constan

Question

Sam heaves a bowling ball with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 45.3 m/s2 for a height 69.0 cm . He releases it at height 2.28 m above the ground. You may ignore air resistance.

QUESTION A

What is the speed of the bowling ball when he releases it?

Answer: v = _____ m/s

QUESTION B

How high above the ground does it go?

Answer: h = _____ m

QUESTION C

How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.86 m above the ground?

Answer: t = _____ s

Explanation / Answer

a) vf^2 = vi^2+2ad

vf^2 = 0+2*45.3*0.69

vf = 7.91 m/s

b) vf^2+vi^2+2gd

0 = 7.91^2-2*9.8*d

d = 3.19 m

h = 3.19+2.28 = 5.47 m

c) d = vi*t+1/2gt^2

0 = 7.91*t-1/2*9.8*t^2

4.9t^2 = 7.91t

t = 1.61 sec

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

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