The player shoots the basketball with an initial velocity of 20 m/s at an angle

Question

The player shoots the basketball with an initial velocity of 20 m/s at an angle of 32 degrees above the horizon.

The player lets go of the ball when the ball is 3 m above the basketball court.

The ball lands in the basket that is 3 meters above the basketball court.

Assume no air resistance and the acceleration in the y-direction is ay = -9.8 m/s2

Answer the following in any order. The origin is at the hand of the player as the ball is released.

The line from the player to the basket is the +x-dir while upward is the +y-dir.

Initial Velocity, Vo = 20 m/s    Initial Angle with horizon, o = 32º

the origin is 3 m above the floor

please find the following:

A. Initial X-velocity

B. Inital Y-velocity

C. Final X-position, xf in meters (just as ball enters the basket)

D.Final Y-velocity, Vfy in m/s (just as ball enters the basket)(it will be negative)

E. Time basketball is in the air in seconds (out-of-hand to entering basket)

Explanation / Answer

Initial Velocity, Vo = 20 m/s    Initial Angle with horizon, o = 32º

A.
Initial X velocity = 20 * cos(32) m/s
Initial X velocity =16.96 m/s

B.
Initial Y velocity = 20 * sin(32) m/s
Initial Y velocity =10.59 m/s

C.

Time ball is in the air = 2*t
t = 10.59/9.8
t =1.08 s
Time ball is in the air = 2*1.08 = 2.16 s

Final X position = Total time * Speed
Final X position = 2.16 * 16.96
Final X position, xf = 36.63 m

D.)
Vfy = 9.8 * 1.08
Vfy = - 10.58 m/s (-ve sign means going downwards)

E.)
Time ball is in the air = 2.16 s

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