# Assume: A 78 g basketball

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Assume: A 78 g basketball is launched at an angle of 44.2◦ and a distance of 11.9 m from the basketball goal. The ball is released at the same height (ten feet) as the basketball goal’s height.
A basketball player tries to make a long jump-shot as described above.
The acceleration of gravity is 9.8 m/s2 . What speed must the player give the ball? Answer in units of m/s
Set the height of 10 feet to be at y=0. Then it becomes a problem of projectile motion for the ball to travel 11.9 m at an launch angle of 44.2 degrees. Let V be the initial speed of launch. $V_x=Vcos\theta$ and $V_y=Vsin\theta$. Define "UP" as the positive y direction.
Based on the vertical motion, one can find how long the ball travels in air $\Delta y=0=V_y\Delta t-\frac {1}{2}g\Delta t^2$.
If the ball hits the goal, it will travel 11.9 m at the time when the ball returns to y=0. Therefore $\Delta x=V_x\Delta t$.