Projectile Motion thrown basketball

  • Thread starter Cate
  • Start date
  • #1
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Homework Statement



A basketball player throws the ball at a 48° angle above the horizontal to a hoop whose centre is located a horizontal distance L = 4.30 m from the point of release and at a height h = 0.50 m above it. What initial speed is required if the basketball is to reach the centre of the hoop? (Ignore the size of the ball.)


Homework Equations



I've tried using two equations:

1.Vo= square root* gh/2sintheta

2. R= Vo^2Sin2theta/g

The Attempt at a Solution




Using the second formula I isolated or Vo and got 6.51 m/s but that wasn't the right answer also, that formula doesn't take into account the height while the first formula doesn't take into account the range...I know this isn't that dificult but if someone could point me in the right direction it would be appreciated!
 

Answers and Replies

  • #2
325
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Okay, first write an algebraic value for the x and y components of initial velocity.

Then using constant acceleration formulae, calculate how long it will take for the ball to reach its maximum height, double that to find how long it takes for it to go up and come back to its initial height.

Again using the formulae, solve for how long it will take the ball to fall a further 0.50m.

Now you know exactly how long the ball is in the air for (t) as a function of Vinitial y. So now you want to write out an equation of Vinitial x for the ball to travel 4.3m in t written out in the first equation.

You should now have two equations you can solve simultaneously and then use the pythagoras theorem to return to the initial velocity not in terms of x and y.
 

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