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Urgent projectile motion

  • Thread starter miley1234
  • Start date
  • #1
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Urgent!!!! projectile motion

Heyy every1
Q Exactly 2.7s after a projectile is fired into the air from the ground ,it is observed to have a velocity v = (8.8i+4.5 j) m/s where the x-axis is horizontal and the y axis is positive upward .
a)Determine the horizontal range of the projectile
b)Determine the maximum height above the ground
c)Determine the speed of motion just above the projectile strikes the ground .
d)Determine the angle of the motion just before the projectile strikes the ground .

I managed to do the first two parts where max height is 49m and range is 56 m .Bt i do not noe how to go abt the c) and d) part
 

Answers and Replies

  • #2
6,054
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What is horizontal component of the velocity? What is the vertical component of final velocity?
 
  • #3
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well i was able to calaculate the max height and range using horizontal velocity to be 8.8 and vertical final velocity as 4.5 at 2.7 s .This much information only was given n the question tht s why i m confused
 
  • #4
CAF123
Gold Member
2,906
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For d), knowing the horizontal and vertical components of initial velocity should help you. You know that the trajectory is parabolic and so just prior to impact, use the condition,[tex] -v_{fy} \hat{y} = v_{oy} \hat{y}. [/tex]
 
  • #5
1,065
10


Horizontal velocity(i component) remains constant since no forces acting on it.
Vertically(j component) there is negative acceleration.
Sketch a graph of velocity vs. time. Upward as positive.
Δv/Δt is constant. You can find initial vertical velocity from acceleration equation.

Speed it the vector sum of vertical and horizontal velocity.

Angle of projection is just the tangent of vertical and horizontal velocity.
 
Last edited:

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