Question

A basketball leaves a player's hands

at a height of 2.20 mm above the floor. The basket is 2.80 mm above the floor. The player likes to shoot the ball at a 40.0 ∘∘ angle. If the shot is made from a horizontal distance of 5.70 mm and must be accurate to ±0.22m (horizontally), what is the range of initial speeds allowed to make the basket?

Answer 1

answer) let the height be h

h=ho+vyt-1/2gt²

h=2.2+vsin40*t-1/2*9.8*t²

at height 2.80 we have

2.80=2.2+vsin40t-4.9t²..............1)

the range is given by

d=vxt

t=d/vcos40

replacing t in eqn 1 we get

2.80=2.2+vsin40*(d/vcos40)-4.9(d/vcos40)²

0.6=dtan40-4.9d²/v²cos40²

v=d/cos40*(4.9/dtan40-0.6)).............2)

the largest range=5.70m+0.22m=5.92m

so from eqn 2 we have

v=5.92/cos40*(4.9/5.92tan40-0.6)=8.19 m/s

for lower range, d=5.70-.22=5.48 m

v=5.48/cos40*(4.9/5.48tan40-0.6)

v=7.92m/s

so answer is

7.92 m/s <v<8.19 m/s or 7.92 m/s <v<8.20 m/s