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Throws a stone horizontally with a velocity of 20ms-1

  • Thread starter moimoi24
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  • #1
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Homework Statement



1. Don Joriel stands on a vertical cliff edge throwing stones into the sea below. He throws a stone horizontally with a velocity of 20ms-1, 560 m above sea level.
a. How long does it take for the stone to hit the water from leaving Don Joriel’s hand?
Use g = 9.81 ms-2 and ignore air resistance?
b. Find the distance of the stone from the base of the cliff when it hits the water.

2. When Maggie applies the brakes of her car, the car slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes?



Homework Equations





The Attempt at a Solution


please help me with those problem, im confused. I am not sure if i got the correct answer. here are my answers 1. a. 2.04 s
b. 20.39 m
2. 18.75 m

Hope you can help me, if got the wrong answer please show me the solution... Thanks a lot...
 

Answers and Replies

  • #2
rl.bhat
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Show your calculations.
 
  • #3
Simon Bridge
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What rl.bhat said: if you don't show us what you tried or your reasoning, then we cannot know how to help you best.
 
  • #4
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wait i'll show you when i got home.. thanks...
 
  • #5
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for 1a. i use v = u + at,
b. s = ut +1/2 at squared

for 2. s = 1/2 (u +v)t

are those correct?
 
  • #6
CAF123
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for 1a. i use v = u + at,
b. s = ut +1/2 at squared

for 2. s = 1/2 (u +v)t

are those correct?
I don't see how you can use the equation in 1a without first knowing the final velocity of the ball (I.e the velocity of the ball prior to impact with the ground). From your answer, it seems you have assumed this to be 0, but this is not the final vertical component of velocity.
 
  • #7
CAF123
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For 1b, we are considering a horizontal distance. There is no acceleration of the ball in x direction so the eqn you were using simplifies to simply [itex] s = v_{ox}t [/itex].
For 2, if I understand the wording of the question correctly, you are correct.
 
  • #8
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I don't see how you can use the equation in 1a without first knowing the final velocity of the ball (I.e the velocity of the ball prior to impact with the ground). From your answer, it seems you have assumed this to be 0, but this is not the final vertical component of velocity.
For 1b, we are considering a horizontal distance. There is no acceleration of the ball in x direction so the eqn you were using simplifies to simply [itex] s = v_{ox}t [/itex].
For 2, if I understand the wording of the question correctly, you are correct.
what is the answer then? please show the solution...
 
  • #9
rl.bhat
Homework Helper
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for 1a. i use v = u + at,
b. s = ut +1/2 at squared

for 2. s = 1/2 (u +v)t

are those correct?
Before you attempt this problem, please go through the projectile motion.
 
  • #10
CAF123
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what is the answer then? please show the solution...
We are not allowed to simply hand over solutions. For 1a, you can use the equation [itex] s_y = v_{oy}t - \frac{1}{2}gt^2 [/itex] right away to find the time of descent. Alternatively, the final vertical component of velocity can be found first so you can then use [itex] v_{fy} = v_{oy} -gt [/itex].
Yet another way is to do all three problems via sketching v-t diagrams for each of the x and y components of velocity.
 
  • #11
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what is the answer on the problem? i am totally confused.. please help me...
 
  • #12
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We are not allowed to simply hand over solutions. For 1a, you can use the equation [itex] s_y = v_{oy}t - \frac{1}{2}gt^2 [/itex] right away to find the time of descent. Alternatively, the final vertical component of velocity can be found first so you can then use [itex] v_{fy} = v_{oy} -gt [/itex].
Yet another way is to do all three problems via sketching v-t diagrams for each of the x and y components of velocity.
i didnt get your equation because of so much symbols, sorry...
 
  • #13
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i got it already... thanks...
 
  • #14
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how about number 2?
 
  • #15
rl.bhat
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Can you find the time taken by a stone dropped from a height 560 m?
 
  • #16
CAF123
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what is the answer on the problem? i am totally confused.. please help me...
Ok, for 1a, initial vertical component of velocity is 0, right? This is because the ball is projected horizontally. So in our equation, [itex] v_{oy} =0. [/itex] The eqn therefore reduces to [itex] s_y = \frac{1}{2}at^2. [/itex] If you define downwards as negative, then the ball will fall downwards 500m so [itex] s_y = -500m. [/itex] Think of the ball moving 500 units down the y axis (I.e starts at y =0 and finishes at y = -500). Now, if downwards is negative, a =-g and so our final equation is [itex] -s_y = -\frac{1}{2}gt^2 [/itex] Rearrange for t and you are done with 1a.
 
  • #17
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so t = 10.68... by the way where did you get 500? is it 560?
 
  • #18
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how about 1b?
 
  • #19
CAF123
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Yes, sorry 560m not 500m. For 1b, again use the relation [itex] s_x = v_{ox}t + \frac{1}{2}at^2, [/itex] but this time considering the x direction. No forces act on the body in the horizontal direction so ax = 0. The horizontal distance covered will be proportional to the time of flight because of a constant (non zero) horizontal component of velocity.
 
  • #20
Simon Bridge
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i got it already... thanks...
Please tell us what you got... be specific: it makes sure we have not accidentally sent you up the wrong tree. When you show working, also be specific: we need to see your reasoning, like how you used an equation, in order to help you properly.

CAF is doing the detail work. Can you see how CAF is choosing which equation to use?
 
  • #21
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is this correct for 1a.
-560 = -1/2 (9.81) t^2
t = 10.68

im not sure if that s correct
 
  • #22
CAF123
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is this correct for 1a.
-560 = -1/2 (9.81) t^2
t = 10.68

im not sure if that s correct
Yes, that is correct. But do you understand why? What makes you unsure?
How did you get on with 1b?
 
  • #23
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tell me if i got 1b correct.
i use your formula,
s = 20 m/s (10.68 s)
s = 213.6 m...

is it right?
thanks a lot....
 
  • #24
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how about the number 2? this is my solution:
s = 1/2(u+v)t
= 18.75 m
 
  • #25
CAF123
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Yes, all questions are correct now. I would now suggest a few more problems to make sure you are on top of things.
 

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