Question

A 2.00-m-tall basketball player is standing on the floor 10.0 m from the basket, as in the figure below. If he shoots the ball at a 40.0° angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m.

Answer 1

I will assume that the center of the basket is 10 m away and 3.05 m high and that the launch height is 2 m

since the angle is at 40 degrees let's look at y(t) first
y(t)=2+v0*sin(40)*t-.5*g*t^2

and x(t)=v0*cos(th)*t

when y(t)=3.05 and x(t)=10, we have a swish as long as the ball is descending.

1.05=v0*sin(40)*t-.5*g*t^2
10=v0*cos(40)*t
t=10/(v0*cos(40))
1.05=10*sin(40)/cos(40)-
.5*g*100/(v0^2*cos^2(40))

v0^2=-
.5*g*100/(cos^2(40)*(1.05-10*tan(40)))

v0=10.67 m/s

check to see if the ball is descending
vy(t)=v0*sin(40)-g*t

it reaches apogee at
v0*sin(40)/g
.7 seconds

at that point x(t)=v0*cos(th)*t
=5.715 meters

so the ball is descending when x(t)=10