# Basic ball projectile motion problem

1. Sep 22, 2011

### miglo

1. The problem statement, all variables and given/known data
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 15 m with the same initial speed of 13 m/s, but in different directions. You through your snowball downward, at 40° below the horizontal; your friend throws her snowball 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.)

2. Relevant equations
V_x=V_0cos(theta)
V_y=V_0sin(theta)
V_f=V_0+at
X_f=X_0+V_0(t)+1/2at^2

3. The attempt at a solution
well this is online and im only given 5 tries to get the correct answer and unfortunately i used up my 5 tries for the first part :(
but anyways i tried finding the time the snowball is 5m from the ground and i used the y=y_0+V_0(t)-1/2gt^2 but im not sure what i should use for the y_0 and V_0, i would say y_0 would be 15 and V_0 would be V_0sin(theta) or 13sin40, then g would be 9.8 and y=5 so then i can solve for t? then after finding t do i just use V_f=V_0+at to get the velocity of the snowball when its 5 m above the ground?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 22, 2011

### miglo

ok so i checked and i only have one chance at getting part 2 of this question correct
so i tried solving for t by using 5=15+8.36t-4.9t^2 which gave me t=2.5
then plugging this value of t into V_f=V_0+at i got V_f=8.36+(-9.8)(2.5) which gave me V_f=-16.14m/s
now to me this kind of makes sense since by then the snowball would be heading downward and its velocity should be negative but im not sure if i plugged in the correct values to the equations i used

3. Sep 22, 2011

### Staff: Mentor

I don't see a part 1 and part 2 in the question statement. I only see a single question: "What is the speed of each ball when it is 5.0 m above the ground?"

What comprises parts 1 and 2?

4. Sep 22, 2011

### Staff: Mentor

Gravitational acceleration applies only to the vertical component of the velocity. So you work out the final y-velocity at 5m above the ground (at time t), then combine with the invariant x-component of the velocity to determine the speed. Be sure to watch out for the signs of things (initial velocities, gravitational acceleration).

You might want to consider using an energy conservation approach for this problem. Then you can deal directly with the speeds and changes of height.

5. Sep 22, 2011

### miglo

part 1 was figuring out the speed of your snowball and part2 was figuring out the speed of your friends snowball
unfortunately i plugged in my answer before anyone could reply to my thread and i got it wrong :( so yeah got both parts wrong, but i still want to know how to solve this problem

gneill i dont understand what you mean by combining with the invariant x-component of the velocity, oh and we haven't gone over conservation of energy, we just started on forces

6. Sep 22, 2011

### Staff: Mentor

For a projectile near the Earth's surface (which assumes a constant acceleration due to gravity), the horizontal component of the motion is independent of the vertical component of the motion. If Vx and Vy are the components of the velocity at some instant, then the speed of the object is $\sqrt{Vx^2 + Vy^2}$.

7. Sep 22, 2011

### miglo

ohhh i see
this whole time i thought i just had to use the y-component for velocity
well thanks a lot gneill