A baseball is seen to pass upward by a window 20 m above the street with a verti

Question

A baseball is seen to pass upward by a window 20 m above the street with a vertical speed of 12 m/s.

A) If the ball was thrown from the street, what was its initial speed?

B) If the ball was thrown from the street, what altitude does it reach?

C)If the ball was thrown from the street, when was it thrown? (That is time elapsed from throwing to reaching the window.)

D)If the ball was thrown from the street, when does it reach the street again? (That is time elapsed from the reaching the window on the way up to reaching the street.)

Explanation / Answer

Given that

finial velocity v=12 m/s

hight h=20 m

now we find the initial speed

v^2-u^2=-2gh

12^2-u^2=-2*9.8*20

u^2=144+392

initial speed u=23.2 m/s

now we find the amplitude does it reach

amplitude H=u^2/2g=23.2^2/19.6=27.5 m

now we find the time to thrown the window

time t=u/g=23.2/9.8=2.4 sec

now we find the time to reach the window

time t=[2h/g]^1/2=[2*20/9.8]^1/2=2.4 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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