You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of $14$ m with the same initial speed of $12$ m/s, but in different directions. You throw your snowball downward, at $40^{\circ}$ below the horizontal; your friend throws her snowball upward, at $40^{\circ}$ above the horizontal. What is the speed of each ball when it is $5.0$ m above the ground? (Neglect air resistance.)

What I did: so $V_0x=$Unknown.

$V_0y= 12$m/s

$\tan(40)=Vy/Vx$ so I got $V_0x=7.713$ m/s

To get the final velocities I used: $V_f^2= V_0^2+2gh$

$V_fx=15.365$ and $V_fy=17.905$.

Then to get total I did $V=\sqrt{17.905^2 + 15.365^2}$ and got $23.59$ m/s but that isn't correct. Can someone help me with what I did wrong.

  • 1
    $\begingroup$ You should probably ask this on physics.stackexchange.com $\endgroup$ – Lucas Apr 16 '15 at 0:29
  • $\begingroup$ Why would $V_{0x}=0$ ? $\endgroup$ – anderstood Apr 16 '15 at 0:55
  • $\begingroup$ I would just assume since they only gave one V0 value $\endgroup$ – Sondra Apr 16 '15 at 1:07

Here is how you go about solving this question. Since the snowballs are being thrown at angles, you need to separate your initial speed into $i$ and $j$ components.

So for the case of Snowball1: $v=12m/s$ at $40$ degrees

  • $u_i = 12\cos(40)$
  • $u_j = 12\sin(40)$

The formula you need is $v^2 = u^2 +2as$

First, find final speed in the $j$ direction. List the variables you know:

  • $v_j$ = what we want to find
  • $u_j = 12\sin(40)$
  • $a_j = 9.8$
  • $s_j = 8$

(we started at distance = $14$ and we want speed at distance $5$ ...letting downwards be positive for simplicity)

Sub these values into our formula and you will get a value for $v_j$.

Next, find final speed in the $i$ direction:

  • $v_i$ = what we want to find
  • $u_i = 12\cos(40)$
  • $a_i = 0$
  • $s_i$ = we don't know but it since $a_i = 0$ this term will cancel out

Sub into our formula and solve for $v_i$.

Final step is to find $v$ by using pythagoras. $(v^2 = v_i^2 + v_j^2)$

My solution for speed of snowball1 at $5m$ above ground = $17.865m/s$

Follow same method for snowball2.

  • $\begingroup$ Welcome to MSE! Here is a guide on how to format your math. $\endgroup$ – mrp Apr 16 '15 at 21:36

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