Projectile motion arrow help

  • #1
8
0
1. An arrow is shot upward at a 30 degree angle compared to horizontal in order to hit a target that is 100 meters away. What was the initials peed of the arrow if it hits the center of the target?


2. This is the hardest question on my sheet, can someone please walk me through how to do it?


3. This stumped me and I really have no idea how to start. Could someone please at least start me on this problem?
 

Answers and Replies

  • #2
The trick to solving these problems are detaching horizontal movement and vertical movement. What I mean by this is thinking about each dimension separately.

Okay, so we first list what we know.
θ = 30 degrees, d = 100 meters, g = -9.8 meters per second squared

We also need these kinematics equations:
d = vi * t + 1/2 * a * t^2
v = vi + a*t

We then think about the problem in the horizontal. There is no acceleration in the horizontal, so we use our first kinematics equation:
d = vh*t

We can rewrite vhorizontal as voriginal * cos(θ). Now, the equation for horizontal distance is:
d = vi * cos(θ) * t.

Now, we have to determine t.

We can't determine the t from the horizontal, so we switch over to thinking in the vertical.

We use the second kinematics equation, using gravity as our acceleration. At the peak of the arrow's path, vv = 0.

The equation is now:
vi * sin(θ) - 9.8 * ttop = 0

However, the fact remains that, at the moment, we are solving for the time it takes to reach the top of the arrow's path. We need the time it takes to hit the ground. We know that it takes the same amount of time to go up as the amount of time to go down.

So, the equation for t is:
t = ttop * 2

ttop = t/2

Now, we plug that into vi * sin(θ) - 9.8 * ttop = 0. Rewrite the equation to solve for t.

Plug the equation for t we just found into the equation for horizontal distance:
d = vi * cos(θ) * t

I trust you can take over from here.

Hope this helped!
 
Last edited:
  • #3
8
0
The trick to solving these problems are detaching horizontal movement and vertical movement. What I mean by this is thinking about each dimension separately.

Okay, so we first list what we know.
θ = 30 degrees, d = 100 meters, g = -9.8 meters per second squared

We also need these kinematics equations:
d = vi * t + 1/2 * a * t^2
v = vi + a*t

We then think about the problem in the horizontal. There is no acceleration in the horizontal, so we use our first kinematics equation:
d = vh*t

We can rewrite vhorizontal as voriginal * cos(θ). Now, the equation for horizontal distance is:
d = vi * cos(θ) * t.

Now, we have to determine t.

We can't determine the t from the horizontal, so we switch over to thinking in the vertical.

We use the second kinematics equation, using gravity as our acceleration. At the peak of the arrow's path, vv = 0.

The equation is now:
vi * sin(θ) - 9.8 * ttop = 0

However, the fact remains that, at the moment, we are solving for the time it takes to reach the top of the arrow's path. We need the time it takes to hit the ground. We know that it takes the same amount of time to go up as the amount of time to go down.

So, the equation for t is:
t = ttop * 2

ttop = t/2

Now, we plug that into vi * sin(θ) - 9.8 * ttop = 0. Rewrite the equation to solve for t.

Plug the equation for t we just found into the equation for horizontal distance:
d = vi * cos(θ) * t

I trust you can take over from here.

Hope this helped!
Hey, thanks! I tried to solve the problem, can you tell me if this is correct?
http://imgur.com/LzbTTlx
 
  • #4
Sorry, but 20 is not the answer.

In fact, there is a much easier way to do this problem, using one formula.

The formula is:

R = vi2*sin(2θ)/g

R is the range, or your horizontal distance. If you check with this, it gives you the value of vi as ~33.6 m/s.

Generally, however, the "long" way of doing the problem is better. This is just a really quick shortcut, but you won't actually learn anything the short way. :D
 
  • #5
haruspex
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Hey, thanks! I tried to solve the problem, can you tell me if this is correct?
http://imgur.com/LzbTTlx
A few problems there.
You've used cos of the angle for both vertical and horizontal. That can't be right.
I don't understand how you get t = 0.177. The time taken will be some seconds.
Please don't post handwritten algebra as images. Make the effort to type it into the post. That bit of effort by the person posting saves effort for everyone reading.
 

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