Question

You both throw 2 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° below the horizontal;the other upward, at 40° above the horizontal. What is the speed of each ball when it is 5.0 m above the ground? (Neglect air resistance)
so Vox=Unknown and Voy= 12m/s
tan(40)=Vy/Vx so i got Vox=7.713m/s
velocity in the x vector doesn't change (no acceleration) so vf=7.713
Vf^2= V0^2+2gh and got Vfy=17.905
To get speed: V=sqrt(17.905^2 + 7.713^2) and got 19.50 m/s but it isnt correct. Any help is appreciated

Answer 1

I feel like the part I might have messed up on what determining what the initial velocity was for, whether it is a y vector or to be used as an hypotenuse to determine the relative x and y vector velocities. I have been working on the question for a long time and any help is appreciated

Answer 2

Ok, first thing I notice is that Voy (initial velocity in the y direction) isn't 12 m/s.

To decompose vectors into the x and y coordinates, we need to make a triangle and use trig functions.



Vi is your initial velocity. In this case, it's 12 m/s, and theta (The O with a belt on it) is your angle (+40 for one snowball, -40 for the other).

Can you find Vx and Vy for your snowballs?

Answer 3

exactly what I needed thank you! and the speeds are the same correct?

Answer 4

Once you get that, your approach should work fine.

Alternatively, we could examine the energy:
\[\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mg\Delta h\]
\[\frac{1}{2}v_f^2 = \frac{1}{2}v_i^2 + g\Delta h\]
\[v_f^2 = v_i^2 + 2g\Delta h\]
Where Delta h is the change in height from the starting point.

You'll notice that this is exactly what your final equation is.

Answer 5

Yep, the speeds will be the same for both snowballs :)

Answer 6

Youre awesome. Thank you again!

Answer 7

My pleasure :)