A 2.00 m tall basketball player wants to make a goal from 10.0 m from the basket

Question

A 2.00 m tall basketball player wants to make a goal from 10.0 m from the basket, which is 3.05m high. If he shoots the bal at a 45 degree angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard?
I have worked on this 3 days, using yahoo answer and other cramster posters. The closest that made sense was an Oracle post, but the answer wasn't finished, and there are so many vi left, I still can't figure it out.
so far...
Vix = vi cos 45
= vi (.707)
Viy = vi sin 45
= vi (.707)
t = delta x/vix
t = 10.0m/ vix

Delta y = 3.05m - 2.00m
Delta y = 1.05m

delta y = viyt + 1/2 at^2
1.05m = vi (.707) (10.0m/vi .707) + 1/2 (-9.8m/s^2) (10.0m/vi .707)^2
1.05m = vi (.707) (10.0m/vi .707) + -4.9 m/s^2 (10.0m /vi /707)^2

????I could follow until this last step. I'm not certain if this isn't the most straight forward way to solve this, or if I'm just having trouble with the last step.

Explanation / Answer

There are two speed component. One is invertical speed and one is horizonal speed let Vx = horizonal speed Vy = vertical speed. Vx = Vcos45 Vy = Vsin45 The displacement of the ball in horizonal direction x = vt 10 = Vcos45 t the time it takes the ball to go throught the basket vertically is also the time it takes the ball to go through the ball horizonally. solve for t t = 10 / Vcos45 the distance placement of the ball in vertical direction Xf = .5at^2 + Vt + Xi we know that t = 10 / Vcos45 3.05 = .5(-9.8) (10 / Vcos45)^2 + Vsin45 (10 / Vcos45) + 2 now just solve: 1.05 = .5(-9.8) (10 / Vcos45)^2 + Vsin45 (10 / Vcos45) 1.05 = -4.9 (10 / Vcos45)^2 + 10 -8,95 = -4.9 (10 / Vcos45)^2 1,826= (10 / Vcos45)^2 /sqrt 1,351 = 10 / Vcos45 1,351 Vcos45 = 10 V = 10 / (1,351 cos45) V = 10,467 m/s so the intial velocity of the ball is about 10,467 m/s

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