Question

1.     If the basket ball hoop is at a height of 3m from the floor at a distance of 4m from the player, what should be the velocity and angle of throw to gain a score?​

Answer 1

Explanation:

The acceleration is that of gravity: 9.80m/s

2

, downward.

(b) The horizontal component of the initial velocity is v

xi

=v

i

cos40.0

0

=0.766v

i

, and the time required for the ball to move 10.0m horizontally is

t=

v

xi

Δx

=

0.766v

i

10.0m

=

v

i

13.1m

At this time, the vertical displacement of the ball must be

Δy=y

f

−y

i

=3.05m−2.00m=1.05m

Thus, Δy=v

yi

t+

2

1

a

y

t

2

becomes

1.05m(



v

i

sin40.0

0

)



v

i

13.1m

+

2

1

(−9.80m/s

2

)

v

i

2

(13.1m)

2

or 1.05m=8.39m−

v

i

2

835m

3

/s

2

which yields,

v

i

=

8.39m−1.05m

835m

3

/s

2

=10.7m/s