You're attending a professional basketball game. the home team calls timeout, and, as the music gets cranked up, the t-shirt crew comes out onto the floor to fire team t-shirts into the crowd. like everyone else, you stand up and wave your arms wildly, hoping that one of the t-shirt guys will fire a shirt your way. the t-shirts are fired from a launcher that uses compressed air. one of the launchers is pointed your way, and is just about to launch a t-shirt at a 45° angle from the horizontal. you are a horizontal distance of 20.0 m from the launcher and a vertical distance of 3.00 m above it. neglect air resistance, and assume that the acceleration due to gravity is 9.80 m/s2.

what is the launch speed needed so that the t-shirt reaches you, rather than one of the hordes of people sitting near you?

The launch speed has to be 15.2 m/s so that the T-shirt reaches you.

Explanation:

The equation for the position of an object moving in a parabolic motion is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity

Please, see the figure for a description of the problem. The vector "r" is the position vector the t-shirt must have to reach you.

As you can see from the figure, the x-component of the vector "r" is 20.0 m and the y-component is 3.00 m. Then, using the equations for the vector "r" we can obtain the time of flight and the initial velocity:

x = x0 + v0 · t · cos α    (since the origin of the frame of reference is located at the launching point, x0 = 0)

x = v0 · t · cos α

Solving for v0:

x/(t · cos α) = v0

20.0 m/ t · cos 45° = v0

Replacing v0 in the equation for the y-component of the vector "r":

y =  y0 + v0 · t · sin α + 1/2 · g · t²       (y0 = 0, same as x0)

y = 20.0 m · sin 45°/cos 45° - 1/2 · 9.80 m/s² · t²

3.00 m = 20. 0 m · tan 45° -  1/2 · 9.80 m/s² · t²

3.00 m - 20. 0 m · tan 45° = -4.90 m/s² · t²

-17.0 m/  -4.90 m/s² = t²

t = 1.86 s

Now, with the time, we can calculate v0:

v0 = 20.0 m/ t · cos 45° = 20.0 m/ 1.86 s · cos 45° = 15.2 m/s

The launch speed has to be 15.2 m/s.

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