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A projectile is thrown into space so as tohave a max. possible horizontal range of 400mts. Taking the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum are? can you show how to solve it?

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Given: there is maximum possible range of 400m.

So, angle of projection = 450.

Now, u2sin2(450)/g = 400

u2= 4000

Now, we know that the velocity of a projectile is minimum at the highest point of its trajectory.

The height of the highest point gives the y co-ordinate of the position of minimum velocity and the half of the total horizintal range gives us the x co-ordinate of that point.

Now, max. height = u2sin2450/2g

  = 4000/2/2g

  = 100m

Thus, the co-ordinates of the point of minimum velocity of the projectile are (200, 100).

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so why it is (200 , 100 ) why not (400 , 100 )
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okkk i understood
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Why Not It is 400,100
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Why not it is (400,100)....?
  • -14
How do we find the angle of projection in this case?
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Why not 400,100
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Because 400 is full range but velocity is minimum at the max. Height and max. Height is in the middle of range
  • -15
Just to think that max range will be at theta=π/4

  • 91
When the horizontal range is maximum, the maximum height attained is R/4 = 100m 
                                      The velocity of projectile is minimum at its highest point.
therefore, Required point is (200,100)
HOPE IT HELPS!!
 
  • 14
As , velocity is minimum at heighest point . So, x- coordinate = 400/2= 200 Again, y= Hmax= u²sin²45°/ 2g ( as range is maximum at 45°) y= u²/4g ( Rmax= u²/g) y= R/4 = 400/4=100 Hence, coordinates are (200,100).
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ohhhh no
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i can't
  • -2
Given: there is maximum possible range of 400m.

So, angle of projection = 450.

Now, u2sin2(450)/g = 400

u2= 4000

Now, we know that the velocity of a projectile is minimum at the highest point of its trajectory.

The height of the highest point gives the y co-ordinate of the position of minimum velocity and the half of the total horizintal range gives us the x co-ordinate of that point.

Now, max. height = u2sin2450/2g

? = 4000/2/2g

? = 100m

Thus, the co-ordinates of the point of minimum velocity of the projectile are (200, 100).
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