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Projectile Motion. Find where it lands.

  • Thread starter TcWhistler
  • Start date
  • #1

Homework Statement


Brett bowls a cricket ball at 144km/h. He releases it at 6 degrees below the horizontal from a height of 2.4m.
How far down the pitch does the ball land?
Here is a bad picture of it. http://img822.imageshack.us/i/physics.png/

Homework Equations



The Attempt at a Solution


I know I have to change km/h to m/s
So, 144km/h = 40m/s

Horizontal Component = 40 cos (6) = 39.72 m/s
Vertical Component = 40 sin (6) = 4.18 m/s
Here is a pic of it all together.
http://img80.imageshack.us/img80/556/physics3.png [Broken]

Now I have no idea where to go from here. Can some guide me on the next step? or If I have done anything wrong.


----------------------------------------------------------------------------------------
Also I do have another question, which isn't on the subject but I think it has a quick answer.

A parcel is dropped from an aeroplane to a boat at sea. The plane is flying at a speed of 100m/s at a fixed altitude of 120m above sea level.
From what time the moment that it is released, does it take for the parcel to hit the water?

http://www.scientificpsychic.com/mind/lean1.html
Theres how to work it out. But I don't understand how we got from
distance = 1/2 × acceleration × time2 + intial speed × time
to
time2 = (2 × distance) / acceleration

How did the 2 x come in?
 
Last edited by a moderator:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
How did the 2 x come in?
It is not 2x, it is two times distance.
In the first problem, using initial velocity as 40sin(6), vertical displacement as 2.4 m, find time to reach the ball to the ground.
Then vcos(6)*t will give you length of the pitch..
 
  • #3
rl.bhat;3138511 In the first problem said:
Thanks, I am still a bit lost. Since I have only started projectile motion

40*Sin(6) = 4.18 ms
Vertical Displacement = 2.4 m

I don't know what equation to use to find the time.
 
  • #4
gneill
Mentor
20,783
2,759
Somewhere in your textbook or notes you will have a general equation of motion that expresses total distance in terms of initial distance, velocity, acceleration, and time. That would be a good place to start. Be sure to take care in assigning directions (signs) to things -- pick a coordinate system and stick with it.
 
  • #5
Somewhere in your textbook or notes you will have a general equation of motion that expresses total distance in terms of initial distance, velocity, acceleration, and time. That would be a good place to start. Be sure to take care in assigning directions (signs) to things -- pick a coordinate system and stick with it.
So far what I found was. V = U at
 
  • #6
gneill
Mentor
20,783
2,759
Keep looking. You should have something that looks similar to:

d(t) = do + vot + (1/2) a t2
 
  • #7
:/ Hmm I can't find anything, I am totally lost..
 
  • #8
gneill
Mentor
20,783
2,759
Well, it'll be there somewhere, perhaps disguised by using different variable names.

For now, take it as given. The equation describes the total distance traveled over time t for a body that has an initial distance do, an initial velocity vo, and undergoing acceleration a.

In your case this would apply to the vertical component of the motion of the cricket ball. The horizontal motion is independent of the vertical motion, and is simply a constant horizontal velocity.

So, given your initial height (do), initial vertical velocity (vo, watch the sign!) and the acceleration due to gravity (a = -g; - because it's directed downwards), solve for the time (t) when the ball hits the ground.
 

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