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Projectile Motion

  1. Sep 15, 2011 #1
    A basketball player shoots a basketball a distance of 6.5 m from the hoop releasing the ball at a height of 1.5 m above the ground. The basketball goal’s rim stands 3 m above the ground. If the player shoots the ball at an angle of 43 degree with the horizontal, what initial speed must he shoot the ball to make it in the basket?

    I know this is a ideal projectile motion problem, but I'm not quite sure where to start. I've tried breaking down the problem into its different compents, but I just can't get that to work

    Typically I'm given a problem with velocity, so this is new to me

    I def. don't won't anyone to give me final answer, but if you could help get me started and walk me through the problem it would be much appreciated
     
  2. jcsd
  3. Sep 15, 2011 #2

    PeterO

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    Homework Helper

    Firstly, you need to be careful that the 6.5 m to the hoop is the horizontal distance, not the direct line - ie it is 6.5m from the point below his arm when he shoots, to the point below the middle of the hoop [we will be considering a point on the middle of the bottom of the ball where he touches it as he shoots, and the middle of the ball will have to end up at the middle of the hoop - "nothing but net".

    The vertical and horizontal components of the balls velocity are Vsin(43) and Vcos(43) - I will leave it to you to determine which one is which.

    Given the vertical component, the ball must go to maximum height and come back down to a point 1.5 m above where it was launched from, in exactly the same time as it travels a horizontal distance of 6.5 m.

    You will get simultaneous equations to solve - and remember the ball will be level with the ring twice - once when it is on its way to maximum height and again when it is coming back down again.
     
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