Question

A 2.0 m tall basketball player is standing a horizontal distance of 7.0 m away from a basketball hoop. The height of the hoop is 3.0 m above the floor. The basketball player throws the ball at an angle of 60 degrees above the horizontal, and from an initial height of 2.0 m above the floor. (Use g= 9.8 m/s2 and neglect air friction effects). Give your answer to 2 significant figures. a) How long does it take for the ball to reach the hoop? b) What is the initial velocity of the ball? c) At what angle to the horizontal is the ball traveling just before it enters the hoop? d) What is the maximum height reached by the ball?

Answer 1

a)

Initial Horizontal and Vertical components of velocities are

V_ox =V_oCos60

V_oy=V_oSin60

The Time taken for the ball to move 7 m horizontally is

From

Y=Y_o+V_oyt-(1/2)gt²

3=2+(V_oSIn60)(7/V_oCos60) -(1/2)(9.8)(7/V_oCos60)²

960.4/V_o²=11.124

V_o=9.29156 m/s

t=7/(9.29Cos60) =1.506743729 s

t=1.5 s (approx)

b)

V_o=9.3 m/s (approx)

c)

V_fx=V_ox=9.3Cos60

V_fy=V_oy-gt =9.3Sin60-9.8*1.5

V_fy =-10.12 m/s

angle

o=tan^-1(-10.12/9.3Cos60)

o=-55.3^o or 55.3^obelow the horizontal

d)

From

V_fy²=V_oy²-2g(H_max-h)

at maximum height V_fy=0

0=(9.3SIn60)²-2*9.81*(H_max-2)

H_max=5.3 m (approx)