Projectile Motion. What initial speed should she give the ball so that it reaches the basket and scores?

Question

An Olympic basketball player shoots towards a basket that is 5.86 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.57 m above the floor at an angle of 43.0o above the horizontal.
What initial speed should she give the ball so that it reaches the basket and hopefully scores?

Answer 1

jmurphy's constituent equations seem correct but there must have been a math error because the solution isn't. From his first equation the flight time is 2.183 s, and from the 2nd equation the ball is 17.9 m below its launch point.
Per ref. 1, the solution is
yt = range*tan(theta) - g*range^2*(1+tan(theta)^2)/(2*v0^2) ==>
v0 = sqrt(0.5g*range^2*(1+tan(theta)^2) / (range*tan(theta)-yt)) = 8.89 m/s
(where yt is the difference between launch and target heights, 1.48 m)
You can use the calculator in ref. 2 to verify this (bearing in mind that g = 9.8 there).

Answer 2

The horizontal motion equation for arrival at the basket is
5.86=v0*cos(43)*t
and the vertical at arrival is
3.05=1.57+v0*sin(43)*t-.5*9.81*t^2

solve for v0

3.67 m/s
j

Answer 3

Covervation of skill. PE + KE before = PE + KE after initially PE = mgh (h is your hand height) very final PE = mgh (h is her heigt in tree) preliminary KE = a million/2mv^2 (might desire to sparkling up for this v) very final KE = a million/2 mv^2 (v given) upload each and every little thing up (be conscious that m cancels) and sparkling up for preliminary v.