Basic ball projectile motion problem

  • Thread starter miglo
  • Start date
  • #1
97
0

Homework Statement


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.)

m/s(your snowball)
m/s(your friend's snowball)


Homework 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


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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
97
0
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
gneill
Mentor
20,925
2,867
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
gneill
Mentor
20,925
2,867
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?

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
97
0
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
gneill
Mentor
20,925
2,867
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 [itex] \sqrt{Vx^2 + Vy^2}[/itex].
 
  • #7
97
0
ohhh i see
this whole time i thought i just had to use the y-component for velocity
well thanks a lot gneill
 

Related Threads on Basic ball projectile motion problem

  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
7
Views
12K
Replies
13
Views
2K
Replies
2
Views
5K
Top