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The question asks: A ball is launched from the origin O at ground level V We've been asked to derive the envelope of trajectories equation which I did fine. Then it says: A vertical wall of height H is placed a horizontal distance D from O. Find the minimum height H=Hmin that will ensure the ball cannot get over the wall. How would I go about doing this? Im guessing as we've been asked to derive the envelope equation we use this? Not sure where to go from here!

john
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  • It the top of the wall lies above the envelope, then the ball cannot get over it. – almagest May 13 '16 at 08:42
  • Thanks @almagest See i thought that so then z<H but where do I go from there? If I insert these variables into the envelope equation I get a quadratic for the velocity not the height H? – john May 13 '16 at 08:51
  • What is your envelope equation? – almagest May 13 '16 at 08:52
  • @almagest z= v^2/2g - gx^2/2v^2 – john May 13 '16 at 08:54
  • Ok. Remember that $v$ is fixed - it is the launch velocity. So what you have is an equation that gives the maximum height $z$ as a function of $x$. So plug in $x=D$ (the distance to the wall) and the resulting value of $z$ gives you the minimum height $H$ for the wall. – almagest May 13 '16 at 09:01
  • so my equation is V^2/2g - gD^2/2V^2 < H ? – john May 13 '16 at 09:06
  • Yes, although you are asked for $H_{min}$ – almagest May 13 '16 at 09:13
  • How does that change it? Sorry I've been staring at this for hours I cannot seem to figure it out! – john May 13 '16 at 09:14
  • It is correct that any $H$ greater than $v^2/2g-gD^2/2v^2$ will block the ball. If the question asked for the minimum $H$ that will block the ball, then the answer is $v^2/2g-gD^2/2v^2$. – almagest May 13 '16 at 09:16
  • I have that... but it then asks me to consider the trajectory with which the ball hits the top of the wall when H=Hmin and find the speed and it gave me 0! – john May 13 '16 at 09:23
  • The speed at which it hits the top of the wall? You can get that by conservation of energy. – almagest May 13 '16 at 09:25
  • I really don't know how to do that :(. Ok thanks for helping! – john May 13 '16 at 09:36
  • Gravitational potential energy + kinetic energy is conserved. So $\frac{1}{2}v^2=gh+\frac{1}{2}u^2$, where $u$ is the speed when it is at height $h$ (and $v$ is the initial speed). – almagest May 13 '16 at 10:11
  • We've never really done that with this kind of problem before, only ever using the trajectory equation or envelope equation Is there another way to compute it? If not I'll do it that way! Thanks – john May 13 '16 at 10:15

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