Typical projectile problem

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  • #1
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Homework Statement


There is a projectile which is launched at point "o" and land at "B".It was thrown with 20m/s with angle of 53 degree.Take g=10 m/s^2 sin theta=0.8 and cos theta=0.6
Find the time at which magnitude of x component of velocity of projectile is twice the magnitude of y component of velocity.
upload_2015-4-16_16-2-5.png

Homework Equations


v sub x =u sub x +a sub x t
u sub x=initial velocity in x-direction i.e u cos theta =16 m/s
v sub x =velocity in x direction at time t.
v sub y =velocity in y direction at time t
v sub y =u sin theta +(- 10 t)

The Attempt at a Solution


v sub x=2 (u sin theta +(- 10 t)
but v sub x=u cos theta (as acceleration in x i.e horizontal direction is zero.
so,u cos theta =2 (u sin theta - 10 t)
putting given values give
t=1 sec.
But in book two answer are given
t=1 sec
and t=2.2 sec.
How I can get t=2.2 s ?





 

Answers and Replies

  • #2
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Actually there would be two cases when horizontal velocity would be 2 times of vertical velocity as shown in your attempt.
Consider the case when projectile reaches highest point, calculate the time for it.
Then see the additional time where once again the horizontal vel. would be equal to 2 times the ver. Vel.
 
  • #3
RUber
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You have to look at the return path, so try the negative.
 
  • #4
SteamKing
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Homework Statement


There is a projectile which is launched at point "o" and land at "B".It was thrown with 20m/s with angle of 53 degree.Take g=10 m/s^2 sin theta=0.8 and cos theta=0.6
Find the time at which magnitude of x component of velocity of projectile is twice the magnitude of y component of velocity.
View attachment 82093

Homework Equations


v sub x =u sub x +a sub x t
u sub x=initial velocity in x-direction i.e u cos theta =16 m/s
v sub x =velocity in x direction at time t.
v sub y =velocity in y direction at time t
v sub y =u sin theta +(- 10 t)

The Attempt at a Solution


v sub x=2 (u sin theta +(- 10 t)
but v sub x=u cos theta (as acceleration in x i.e horizontal direction is zero.
so,u cos theta =2 (u sin theta - 10 t)
putting given values give
t=1 sec.
But in book two answer are given
t=1 sec
and t=2.2 sec.
How I can get t=2.2 s ?

Find out how much time it takes for the ball to reach the top of the parabola and start falling back to earth.

In your equation:

vx = 2[u sin (θ) ± 10t],

you picked the negative sign and worked out t. What if you then picked the positive sign and solved this equation? What value of t would you get then?
 
  • #5
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What if you then picked the positive sign
Why?acceleration due to gravity is always pointing downward and I have chosen downward direction to be negative.
 
  • #6
SteamKing
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Why?acceleration due to gravity is always pointing downward and I have chosen downward direction to be negative.
Well, then why did you write the equation with ± ? o_O

The first comment still applies. After the ball reaches its maximum height and starts falling again, there will be another point where the horizontal velocity is twice the verical velocity. :wink:
 
  • #7
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83
Well, then why did you write the equation with ± ?
No.I have not.I was just intended to write v=u+at as a is negative,it became v=u +(-a)t
 
  • #8
2,486
83
The first comment still applies. After the ball reaches its maximum height and starts falling again, there will be another point where the horizontal velocity is twice the verical velocity.
I got your point.As there is symmetry ,there will be another point where the horizontal velocity is twice the vertical velocity.Thanks .:smile:
 

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