# Projectile Motion that ends before max. distance

## Homework Statement

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

## Homework Equations

v^2=iv^2+2ad (i: initial, v: velocity, d: displacement, a: acceleration)
d=ivt+.5a(t^2) (t: time)
a=(v-iv)/t
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.

## Answers and Replies

Welcome to PF.
first find the speed angle with horizontal when ball reaches the height of 2.6 meter then apply the formula of maximum distance. It should work.
Do you know why i say to find the velocity and angle at the time when it is at height of 2.6 meter.(see proof of maximum distance of projectile)