# Basketball Projectile Motion problem

• PHYSICSownsME
In summary, the question asks how fast the ball must leave your hand to pass through the basketball hoop at a distance of 4.92m and a height of 0.457m above the release point, with a launch angle of 30 degrees and a gravitational acceleration of 10m/s2. To solve this, the x and y components of velocity must be calculated using the given equations. The time it takes for the ball to reach the top and come back down can be found by adding the two displacements together.

## Homework Statement

You throw a basketball towards a basketball hoop that is 4.92 m away from you, and 0.457 above the release point at 30 degrees above the horizontal (assume g = 10 m/s2)

How fast does the ball have to leave your hand to pass through the hoop on its way down?

## Homework Equations

So dx = 4.92 m, and dy= 0.457 m
Vx(t) = (Initial velocity x cos30) t
Vy(t) = (Initial velocity x sin30) x gt

## The Attempt at a Solution

I know that I just need to find either the x component or y component of velocity as I have θ = 30° which I can use to find the velocity from the release point.

I feel like this question is missing something. I just can't find the time it takes to reach the top (and then it will come back down and go into the hoop) or time it takes to reach the hoop.

Any help would be appreciated.

You throw a basketball towards a basketball hoop that is 4.92 m away from you, and 0.457 above the release point at 30 degrees above the horizontal (assume g = 10 m/s2)
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The time for the ball to reach 4.92m away is equal to the time it needed to reach 0.457m above.
It is the addition of 2 vectors(displacements) that determine the position of the ball.