A basketball player is standing on the floor 10.0m from thebasket. The height of

Question

A basketball player is standing on the floor 10.0m from thebasket. The height of the basket is 3.05m and he shoots the ballfrom a height of 2.00m and at a 40.0 degree angle with thehoriztonal. At what speed must the player throw the ball so theball goes through the basket without striking the backboard. I started by using the general range equation and put 10.0m asthe range. After that i plugged in everything and got 10.0m =0.05vi^2 + 0.50vi. Then that turns into a quadratic and I got 10m/s for the answer. I think its wrong because I didn't even use the3.05m height of the basket. What did I do wrong here? Any helpwould be appriciated thanks. A basketball player is standing on the floor 10.0m from thebasket. The height of the basket is 3.05m and he shoots the ballfrom a height of 2.00m and at a 40.0 degree angle with thehoriztonal. At what speed must the player throw the ball so theball goes through the basket without striking the backboard. I started by using the general range equation and put 10.0m asthe range. After that i plugged in everything and got 10.0m =0.05vi^2 + 0.50vi. Then that turns into a quadratic and I got 10m/s for the answer. I think its wrong because I didn't even use the3.05m height of the basket. What did I do wrong here? Any helpwould be appriciated thanks.

Explanation / Answer

x=10m, hf=3.05m ,ho=2.00m and =40deg. Vo=? First x=VoCost, and we also know thaty=voSint-1/2gt2, so Vo=10/(C os40t)) then3.05-2.00=(10/Cos40t)Sin40t-4.9t2 or1.05=10Tan40-4.9t2 then t2=1.498 and t=1.224seconds. Now we can easily find the initial velocity from theequation Vo=10/(Cos40(1.224))=10.66 m/s

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