A basketball player shoots a ball, and 1.91 seconds later the ball falls into th

Question

A basketball player shoots a ball, and 1.91 seconds later the ball falls into the net at an angle of 60 degrees from the horizontal. The ball enters without hitting the backboard. The ball was released a horizontal distance of 10 m from the basket. In this problem, treat up as a positive and down as negative. ASsume air resistance is negligible.

A. What is the X component of the ball's velocity at the moment it enters the net?

B. What is the y component of the ball's velocity at the moment it enters the net? Indicate the direction with a plus or minus sign.

C. What is the y component of the ball's initial velocity(@ t=0)? Again indicate with a plus of minus sign.

D. What are the magnitude and angle (with respect to the horizontal) of the ball's initial velocity? Express the angle with two significant figures.

E. If the height of the basket is 3.05 meters, what is the initital height of the ball? Basically, at what height was the ball released.

Explanation / Answer

a) vcos60
s=vcos60*t
v=10.47

b)-vsin60= -9.06

c) ucos=5.23   

-9.06=usin-g*1.91

=61.56, u=10.98

u vertical= usin=9.65

d)61.56 with horizontal

e)3.05-0.7425=2.3m

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