Projectile basketball question

[matti0827]

Homework Statement

A regulation basketball net is 3.05 m above the floor. The three point
circle is about 4.6 metres from the basket- the player 5.0 m. At what speed should
the player shoot if the ball is released at a height of 2.25 m at 45 above the horizontal
to score three points?

dy= 3.05m - 2.25m = +0.8m
dx= 5.0 m
a=g=-9.8 m/s²
angle= 45

Homework Equations

∆dy= v1ysin45 ∆t + 1/2 g∆t²

∆dx= v1xcos45 ∆t

The Attempt at a Solution

do I not need time?

 

[Nate810]

∆dy= v1ysin45 ∆t + 1/2 g∆t2 0.8m = v1sin45∆t + 4.9∆t2v1sin45∆t = 0.8 - 4.9∆t2divide both sides by sin45v1∆t = (0.8 - 4.9∆t2)/ sin45∆dx= v1xcos45 ∆t 5.0m = v1cos45∆tdivide both sides by cos45v1∆t = 5.0/cos45equate the two equations(0.8 - 4.9∆t2)/ sin45 = 5.0/cos450.8sin45 - 4.9sin45∆t2 = 5.0cos45multiply both sides by sin450.8sin45sin45 - 4.9sin45sin45∆t2 = 5.0cos45sin45 0.8sin2 45 - 4.9sin2 45∆t2 = 5.0sin2 45subtract 0.8sin2 45 from both sides- 4.9sin2 45∆t2 = 5.0sin2 45 - 0.8sin2 45divide both sides by - 4.9sin2 45∆t2 = (5.0sin2 45 - 0.8sin2 45)/- 4.9sin2 45 ∆t2 = 0.5077∆t = 0.7126v1∆t = (0.8 - 4.9∆t2)/ sin45v1∆t = 7.2v1 = 7.2/∆tv1 = 7.2/0.7126 v1 = 10.09