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.
Best Answer
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
The formula you need is $v^2 = u^2 +2as$
First, find final speed in the $j$ direction. List the variables you know:
(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:
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.