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Kinematics: Traveling in an arc

  • Thread starter zesi
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  • #1
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Homework Statement


Consider a cannonball being fired with a velocity of 30m/s from a cliff of height 45m.

http://imgur.com/OCMNEPv [Broken]

(a) Calculate the time taken for the ball to reach the ground.

(b) Calculate the range of the motion.

(c) Calculate the horizontal velocity, vertical velocity, and the velocity of the cannon ball as it touches the ground.

Homework Equations



4 kinematics formula.

The Attempt at a Solution



(a) using s=vt+0.5at^2
I found out t=3.0s


I am having a hard time with (b) and (c) because it is travelling in an arc. I am not sure how I should proceed.

(b) I don't think I can assume the calculation that it travels in a straight line down, then use trigonometry to find the range. What other ways can I think of?

(c) If I use "v(f) = v(i) + at" then it will the final velocity will be 60m/s, assuming acceleration (due to gravity) = 10m/s^2. However is this 60m/s the vertical velocity or the velocity as it touches the ground?

From my understanding v(f) = 60m/s is the vertical velocity. Once I have the range (horizontal distance) I can use it to find the horizontal velocity. Using vector concept, taking
vertical velocity - horizontal velocity
= velocity as it touches the ground.

I am having a hard time figuring out the horizontal distance.

Correct me if I am wrong.
Which concept am I not getting it? Hope someone can help me. Thank you.
 
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Answers and Replies

  • #2
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Have you dealt with projectile motion questions in the past?
 
  • #3
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Hi PWiz. No I have not learnt projectile motion. How should I understand the concept?

Thanks
 
  • #4
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You will first have to understand that velocity in two dimensional motion can always be resolved into it's x and y components. If I throw a ball with an initial velocity ##u## into the air at an acute angle theta from the surface, the ball takes an approximately curved (parabolic) path. Can you give me expressions for the two components(x and y) of the velocity in terms of ##u## and theta right at the beginning?
 
  • #5
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Ohhh! I keep thinking because it is an arc and didn't break down into it's horizontal and vertical direction. Here is what I did. Please correct me if I am wrong.

I understand that horizontal distance is constant = 30m/s

So I know the time taken is 3.0sec. 30x3 = 90m.
Therefore, horizontal distance = 30m

Now I can I use v(f) = v(i) + at to find my vertical velocity. Since v(i) = 0m/s
I get 30m/s.

Using vector concept and pythagoras theorem,
final velocity as it touches the ground
= sqrt[(horizontal velocity)^2 + (veritcal velocity)^2]
approximately 42m/s.

I feel makes sense. Am I wrong in ny concept? Thank you.
 
  • #6
694
114
I understand that horizontal distance is constant = 30m/s
I think you mean the magnitude of the horizontal velocity is 30 m/s .
horizontal distance = 30m
I take it that this is also a typo; you've correctly evaluated the distance to be approximately 90m previously.

Other than these two things, your answer (and understanding) is perfectly fine.
 
  • #7
10
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Oh so sorry! Yes I got.

magnitude of Horizontal velocity= 30m/s.

Horizontal distance = 90m

Sorry made a very bad typo.

Thank you for the confirmation PWiz
 

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