Solving for v(f) using trig functions.

  • #1
Apollinaria
82
0
Hi everyone, a classmate and I are studying for a test and have been trying to work out the following problem for the past hour and a half with absolutely no progress. Please point us in the right direction :)

Homework Statement



Someone at a third floor window (12m above ground) hurls a ball downward at v=25m/s and a 45 deg angle. How fast will this ball be traveling when it strikes the ground?

Homework Equations



d = v(i)t + 0.5at^2
D(y) = -12m
A(y)= -9.81m/s
Vi(y)= 0

The Attempt at a Solution



Our first attempt at a solution was to find the time. Then we didn't know where to go from there.

(2 x -12)/-9.81=t^2= take the root of whatever that was and get...... 1.5641s
 

Answers and Replies

  • #2
szynkasz
115
2
You need to calculate velocity at that time:
[tex]v_x(t)=v_x(0)=-v\cos 45^o\\
v_y(t)=v_y(0)-gt=-v\sin 45^o-gt
[/tex]
 
  • #3
Apollinaria
82
0
You need to calculate velocity at that time:
[tex]v_x(t)=v_x(0)=-v\cos 45^o\\
v_y(t)=v_y(0)-gt=-v\sin 45^o-gt
[/tex]

Still confused. Was our time calculated correctly? The 1.56s
 
  • #4
Redbelly98
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You don't actually need to calculate the time. There are some other equations that also deal with constant acceleration -- use the one that does not contain time in it.

Also, you said that vi(y) is zero. What does the problem statement say about the initial velocity?
 
  • #5
Apollinaria
82
0
You don't actually need to calculate the time. There are some other equations that also deal with constant acceleration -- use the one that does not contain time in it.

Also, you said that vi(y) is zero. What does the problem statement say about the initial velocity?

I'm not sure we we're solving it right to begin with. In other problems like these, but without the degree portion, Vi(y) would be zero. Whereas Vi(x) would be the given velocity for sure.
Do not know how to approach this or what values we actually have anymore.
Initially we assumed that...

Vi(x)= 25m/s
D(x)= ?
a(x)= 0m/s2
t= ?

Vi(y)= 0m/s
D(x)= -12m
a(x)= -9.81m/s2
t= ?

Not sure how to use trig functions to determine any other values if the values above are wrong.
 
  • #6
Redbelly98
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In other problems like these, but without the degree portion, Vi(y) would be zero. Whereas Vi(x) would be the given velocity for sure.
If you look at those other problems, they probably say that the object was thrown horizontally, which means in the x-direction.

This problem is different. The problem statement says the object is thrown at a 45 degree angle, downward, at 25 m/s. So you need to draw a vector angled downward at 45 degrees. It should look something like this:

http://flylib.com/books/2/760/1/html/2/FILES/03fig18.gif [Broken]

Except that your vector is 25 m/s, not 33 kph. Then use trig, OR the fact that you can make a 45-45-90 right triangle, to figure out the vertical (y) component of the vector.
 
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  • #7
Apollinaria
82
0
Redbelly, we did this but it doesn't equal the final velocity. Are there more steps to take after this?

25sin45= 17.6776 (Vertical velocity?)

The answer were looking for is something like 29m/s
 
  • #8
HallsofIvy
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At 45 degrees sine and cosine are equal. vertical and horizontal (initial) velocities are the same. When the ball hits the ground, vertical velocity is still the same but vertical velocity will have increased. Calculate the speed of the ball from that.
 
  • #9
Apollinaria
82
0
At 45 degrees sine and cosine are equal. vertical and horizontal (initial) velocities are the same. When the ball hits the ground, vertical velocity is still the same but vertical velocity will have increased. Calculate the speed of the ball from that.

They are. But do you mean that when the ball hits the ground, vertical or horizontal velocity will have increased?
 
  • #10
azizlwl
1,065
10
The easiest way to do is by using conservation of energy.
Here take consideration of the conservative force of gravity
It doesn't mattter what angle you throw, only the height counts.
With single equation of energy you can solve the problem since the question goes for the speed.
 
  • #11
Apollinaria
82
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The easiest way to do is by using conservation of energy.
Here take consideration of the conservative force of gravity
It doesn't mattter what angle you throw, only the height counts.
With single equation of energy you can solve the problem since the question goes for the speed.

I think I know what you're referring to but I will not get any of the 10 marks if I use the energy equation as this is a kinematics unit and he is testing our understanding of kinematics.
 
  • #12
azizlwl
1,065
10
1. Get initial horizontal and vertical velocity.
2. Get final vertical velocity after travelling 12m with acceleration g. Horizontal velocity remain constant(assume no forces acting on it, example air resistance).
3. Use Pythagoras's theorem to find resultant of final vertical and horizontal velocity.
 
Last edited:
  • #13
Apollinaria
82
0
1. Get initial horizontal and vertical velocity.
2. Get final vertical velocity after travelling 12m with acceleration g. Horizontal velocity remain constant.
3. Use Pythagoras's theorem to find resultant of final vertical and horizontal velocity.

OH MY GOD! FINALLY! Amazing. It worked out :)))) Thank you! That was so clever.
 

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