Answer to Question #145184 in Classical Mechanics for Isurini Sirisena

Question

A basketball is thrown with a velocity of 6.0 m s−1 at an angle of
40° to the vertical, towards the hoop.
(a) If the hoop is 0.90 m above the point of release, will the ball
rise high enough to go in the hoop?
(b) If the center of the hoop is 3.00 m away, horizontally, from
the point of release, explain whether or not you believe
this throw will score in the hoop. Support your explanation
with calculations.

Answer 1

a) Yes.

$$H=\frac{v^2\sin^2{40}}{2g}=\frac{6^2\sin^2{40}}{2(9.8)}=1.0\ m>0.9\ m$$

b) No. It cannot achieve this height and this range with initial conditions given.

$$h=0.25R\tan{\theta}=0.25(3)\tan{40}\\h=0.63\ m<0.9\ m$$