text: # Projectiles - golf ball
  
  1. Jan 22, 2004
  
  ### Ihsahn
  
  Ok, having studied projectiles over the past few weeks, reaching the "Equation for the path of a projectile stage," I've been presented with the following question. As far as I can see, there isn't enough info to answer the question (unless, of course, its one of those simultaneous equation ones, or I've missed something obvious)
  
  A golf ball is struck from a point O, leaving O with a speed of 21 root 2 m/s. The ball lands, without previously bouncing, at a point A, which is 10 metres below the horizontal plane through O. Find the speed of the ball at the instant it lands at A
  
  In the question, g (Gravity) is taken to be equal to 9.8m/s^2...
  
  Any help would be much appreciated!
  
  Last edited: Jan 22, 2004
  2. Jan 22, 2004
  
  ### Staff: Mentor
  
  There are several ways to solve this. If you've covered conservation of energy, use it.
  
  Assuming you haven't, then considered the x and y components of the velocity separately. And realize that $V^2=V_x^2+V_y^2$.
  
  3. Jan 22, 2004
  
  ### Julian Solos
  
  Re: Projectiles
  
  Excuse me, but I'm not familiar with the expression 21 root 2. What is it?
  
  4. Jan 22, 2004
  
  ### Ihsahn
  
  Sorry if that was unclear, I meant 21 multiplied by the square root of 2.
  
  Doc Al, I don't quite follow... we haven't covered conservation of energy... and I'm unsure how to obtain the x and y components given that we have no angle of projection? Forgive me if its something obvious or simple.
  
  Thanks.
  
  5. Jan 22, 2004
  
  ### Staff: Mentor
  
  You don't have to know the angle. Call the components $V_x$ and $V_y$. Now write down the final values of those velocity components. Try it and see what you get. (Hint: Use $V_f^2=V_i^2+2as$.)
  
  6. Jan 22, 2004
  
  ### Ihsahn
  
  Again, sorry if this sounds stupid;
  
  We've always evaluated (for example) $V_x$ as $V cos$ a and $V_y$ as $V sin$ a ??
  
  The only thing I can think of from your suggestion is the use of trig identities? Do they play a part?
  
  I'm obviously missing something...
  
  7. Jan 22, 2004
  
  ### Staff: Mentor
  
  No trig needed. Do you agree that $V^2=V_x^2+V_y^2$?
  
  So do the following...
  
  1) Find $V_{y,f}^2$ in terms of $V_{y,i}$, $a$, and $\Delta y$ using the kinematic equation given earlier.
  
  2) Realize that $V_x^2$ is a constant.
  
  3) Add these to get $V_f^2$.
  
  8. Jan 22, 2004
  
  ### Julian Solos
  
  A solution (I think )follows. If you don't want to see it, ignore this post.
  
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  .
  
  Denote the initial velocity of the golf ball $$V_{0}$$ and the final velocity $$V$$.
  
  We have a relation
  
  $$V_{0}^2 = V_{x0}^2 + V_{y0}^2$$.
  
  Substituting $$V_{0} = 21 \sqrt {2}$$,
  
  we have
  
  $$882 = V_{x0}^2 + V_{y0}^2$$.
  
  Therefore,
  
  $$V_{y0}^2 = 882 - V_{x0}^2$$.
  
  Ignoring the effect of friction, the $$x$$ component of the speed of the golf ball remains unchanged throughout its flight.
  
  Thus,
  
  $$V_{x0} = V_{x}$$.
  
  The $$y$$ component of the final speed of the golf ball upon landing is obtained from the following relation:
  
  $$V_{y}^2 = V_{y0}^2 + 2g (10 \mbox{m})$$
  $$= (882 - V_{x0}^2) + 196$$
  $$= 1078 - V_{x0}^2$$
  $$= 1078 - V_{x}^2$$.
  
  $$V^2 = V_{x}^2 + V_{y}^2$$
  $$= V_{x}^2 + (1078 - V_{x}^2)$$
  $$= 1078$$.
  
  $$V = \sqrt {1078}$$
  $$= 32.83 \ \mbox{m/s}$$
  
  Last edited: Jan 25, 2004