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Projectile:Launching and landing at different heights

  • Thread starter uaeXuae
  • Start date
  • #1
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Homework Statement



1.If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what will be the upward (vertical) velocity at the instant of release? What will be the forward (horizontal) velocity?
How high (above the point of release) will the shot go? What is the time it takes the shot to reach its maximum height?

2.If the shot in the problem above is released from a height of 6 ft and later lands on the ground (height = 0.0 ft),
what was the total time of flight? How far did the shot travel horizontally?


Homework Equations



vx=vcos41 = 27.16 ft/s
vy = vsin41 = 23.6 ft/s
dy= (vfy^2-viy^2)/2a) => (0-23.61^2)/[2*(-32)] = 8.71m
vfy = viy + at => 0 = 23.6 + (-32)t => t = 0.7375s

The Attempt at a Solution



for the second part im not sure how to proceed do i use the following equation:
dy= viy^2 + 1/2at^2
6 = 23.6t -16t^t
olving for the quadratic equation i get two value for t
t1 = 1.14848205556934
t2 = 0.32651794443066

By common sense when the height was zero and if i was asked to calculate the total time it would have been 0.7375*2 = 1.475 s. So in part 2 the time should be greater than 1.475s since the height is included so can anyone put me on the correct track please ?
 

Answers and Replies

  • #2
54
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I wrote the quetion and where ive reached in a word file heres a print screen :


http://aycu09.webshots.com/image/34208/2002423633968439027_rs.jpg
 
  • #3
rl.bhat
Homework Helper
4,433
7
6 = 23.6t -16t^t
Here displacement and acceleration are in the same direction and initial velocity is in the opposite direction. So the equation becoms 6 = -23.6t + 16t^2. Now try
 
  • #4
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Thanks again..

i used the following URL to quick solve for the values, there is a real and imajinary value :
URL is :
http://www.webinfocentral.com/_CALC/Calc_Equations.aspx [Broken]

http://aycu09.webshots.com/image/35168/2005967840289974889_rs.jpg


do i take (real^2 + Imajinary^2) ^ 0.5 for both values ?
 
Last edited by a moderator:
  • #5
rl.bhat
Homework Helper
4,433
7
You won't get imaginary values. Use this equation: t = {-b + or -[b^2 - 4ac]^1/2}/2a
 
  • #6
54
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6 = 23.6t -16t^t
Here displacement and acceleration are in the same direction and initial velocity is in the opposite direction. So the equation becoms 6 = -23.6t + 16t^2. Now try
to be sure that we are on the same page i can see that youve used the same equation i did.

d= viyt + 1/2at^2

6 = 23.6t -16t^2

Rearranging the equation =>

16t^2 -23.6t +6 = 0


can u explain why you have it to -6 and not +6 ?
 
  • #7
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Trying both equations;

16t^2 -23.6t +6 = 0

and

16t^2 -23.6t -6 = 0

will still give me time < 0.7375*2

can anyone gimme a hint ?
 
  • #8
rl.bhat
Homework Helper
4,433
7
16t^2 -23.6t -6 = 0

t = {-(-23.6) + or - [23.6^2 - (-4*16*6)]^1/2}/2*16 solve this and take + value because the time cannot be negative.
 
  • #9
54
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16t^2 -23.6t -6 = 0

a=16
b=-23.6
c=-6

Delta= b^2-4ac
Delta =(23.6)^2-(4*16*-6)
Delta = 940.96
sqrt(delta) = 30.67

t1= [-b+sqrt(delta)]/2a
t1= (-23.6+30.67)/(16*2)
t1=0.22s

t2= -b -sqrt(delta)2a
t2= (-23.6-30.67)/(16*2)
t2 = 1.695s

using common sense i should take the second one.
 
  • #10
54
0
6 = 23.6t -16t^t
Here displacement and acceleration are in the same direction and initial velocity is in the opposite direction. So the equation becoms 6 = -23.6t + 16t^2. Now try


Can you explain to me how did you relate

Displacement , Acceleration and Initial velcoity together ? to me i wouldnt have solved it if it came up in a test. So what is the difference between part 2 and part 1 ; why did you take a = 32 and not -32 as i did in part 1.
 
  • #11
rl.bhat
Homework Helper
4,433
7
t1= [-(-b)+sqrt(delta)]/2a
t1= (+23.6+30.67)/(16*2)
t1=1.695s

t2= -b -sqrt(delta)2a
t2= (23.6-30.67)/(16*2)
t2 = -.0.22 it is not possible.
In the first part initial velocity and displacement are upwards and acceleration in the downwards. Out of three values mejority should be positive. In the second case displacement and acceleration are downwards and initial velocity is upwards. Hence the signs are applied.
 
  • #12
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So this concept applies for all problems that have different launching and landing wheras a is considered -32 or -9.8 for problems that have the same launching and landing(height =0)

Question in part 1 is to find the upward so a = -32.
in the 2nd part to find the total time and distance meaning a is pointing downward = + 32

I drew this quickly:

http://aycu33.webshots.com/image/34792/2003791767218454541_rs.jpg
 
Last edited:
  • #13
rl.bhat
Homework Helper
4,433
7
Yes. Nice drawing. Keep it up.
 
  • #14
54
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Thanks man. you were more than helpful to me.
 

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