A 0.3kg baseball moving at 48m/s horizontally is struck by a bat in a \'pop fly\

Question

A 0.3kg baseball moving at 48m/s horizontally is struck by a bat in a 'pop fly': the baseball emerges from the collision moving straight up. If the ball rises 64m above the point where it was struck, what is the magnitude of the impulse that acts on the ball during the collision with the bat?

I don't undertand what happens to use energy to find the velocity after the collision.

initial momentum p=0.3*48i=14.4 i

vy=sqrt(2gh) <-- where does this come from?
=35.435 m/s
final momentum=10.63 j

Impulse= ,change in momentum=10.63j-14.4i
magnitude=17.9 kgm/s

Explanation / Answer

M1 x V1 = P1 = 0.3 x 48=14.4 kgm/s 2as=v^2 - u^2 ,putting the values, a=18 m/s2 v=u+at ,t= 2.67 sec Hence,F=ma=5.4 N and t=2.67 s Impulse = F x t=14.42 Ns

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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