1.60-m-tall girl throws a football at an angle of 41.0 degrees from the horizontal and at an initial velocity of 9.40 m/s. How far away from the girl will it land? Rephrase this question

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initial velocity in x-direction is: 9.40 m/s * cos(41°) ~= 7.094 m/s

initial velocity in y-direction is: 9.40 m/s * sin(41°) ~= 6.167 m/s

if the point (0|0) is set at the girl's feet then

the x-coordinate of the ball is

x(t) = 7.094 * t

and the y-coordinate of the ball is

y(t) = -1/2 * g * t² + 6.167 * t + 1.60 = -4.905 * t² + 6.167 * t + 1.60

The ball landing means y is going to be zero:

0 = -4.905 * t² + 6.167 * t + 1.60

solving this equation gives one positive solution for t:

t ~= 1.478

Now we get

x(1.645) = 7.094 * 1.478 = 10.48m

So the ball will land approximately 10.5m from the girl.