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a 2.00 m tall basketball player is standing on the floor 10 m from the basket. if she shoots the ball at a 40 degree angle with the horizontal at what initial speed mjust he throw the basketball so that it goes through the hoop without striking the backboard? the height of the basket is 3.05 m

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    1.) First, solve in the y direction. You know the ball starts at 2m and ends at 3.05m. You also know that acceleration is -9.8m/s^2, and that initial velocity in the y direction is going to be the overall initial velocity times sin45. You then use the equation x-x0=v0t+.5at^2, which after plugging in your values is 1.05=v0sin(45)t-4.9t^2. Now you just need something to plug in for t, and to do this, create an equation in the x direction.

    2.) In the x direction, acceleration is going to be zero, since the problem does not tell you about any air resistance. Distance is 10m, and the initial velocity in the x direction is going to be the overall initial velocity times cos45. So using the same equation as in 1.) except plugging in numbers from the x direction, you get 10=v0cos(45)t, with no acceleration. You can then see that t=10/(v0cos45).

    3.) Plug this value for t back into the equation in the y direction. 1.05=10v0sin(45)/v0cos(45)-490/(v0cos45)… This simplifies to 8.95=490/(v0cos45)^2, and solving for v0 would give you v0 = (490/(8.95cos^2(45))^(1/2).

    So your final answer is v0=10.46m/s.

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