A basketball leaves a player's hands at height of 2.1 m above floor. the basket is 2.6 m above the floor, player shoots the ball at a 38° angle from horizontal. if the shot is made from a distance of 11.22 meters what is initial speed?
answer is 11 m/s
v^2=iv^2+2ad (i: initial, v: velocity, d: displacement, a: acceleration)
d=ivt+.5a(t^2) (t: time)
max distance=(iv^2 sin2θ)/9.8
max height=(iv^2 sin^2(θ))/(2*9.8)
The Attempt at a Solution
i have tried using 11m as max distance just in case, but did not get desired answer.
thought a lot about using max height, but there are two variables.
i feel like there is just one piece of data missing, but unfortunately there is not.
also, i can get the rates at which some variables are equal to eachother, but i do not see it helping in any manner.
i have literally been unable to do this for nearly 1 and a half hours.