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While solving problems regarding projectiles cant we resolve the acceleration due to gravity along(or against)the directionof initial velocity and use this in the equations of motion to derive the answer?

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That could be done but it will make the question more complex since now you to solve 2 equation 1) in direction of motion other perpendicular to it

Hope it helps

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  • $\begingroup$ A projectile shot at an angle 60 degrees to the horizontal strikes a wall 30 m away at a point 15 m above the point of projection.( g= 10m/sec ^2) . $\endgroup$ – Chappy Mar 31 '17 at 14:34
  • $\begingroup$ What do we have to find? $\endgroup$ – Utkarsh futous Mar 31 '17 at 14:38
  • $\begingroup$ A)Find the speed of projection B)magnitude ofor velocity when it strikes the wall.The traditional method gives the right answer but resolving g along the initial velocity doesn't. I was wondering why. $\endgroup$ – Chappy Mar 31 '17 at 14:47
  • $\begingroup$ Why don't you show your work as it is of course bound to get long by the way you desire or just check again there must be some calculation mistakes $\endgroup$ – Utkarsh futous Mar 31 '17 at 15:01
  • $\begingroup$ UCos60t=30. Which gives ut=60. Also 15=usin60t-.5x10xt^2. Which yields u=22.06m/secand t=2.71 sec.U(x)=22.06Cos60=11.03 m/sec.U(y)=22.06Sin60=19.1m/sec.V(x)=U(x)=11.03.V(y)=19.1-10x2.71=8.Squaring both adding and taking the square root gives13.6 $\endgroup$ – Chappy Mar 31 '17 at 16:13
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You could do that, but you would need to write equations of motion using coordinates which are not horizontal and vertical coordinates. You would have a zero initial velocity in the coordinate perpendicular to the initial velocity, but you would have accelerations in both coordinates. The coordinate directions would be tilted compared to gravity.

The final answers would be in that tilted coordinate system. You could then rotate them back to a horizontal system. But that's adding a lot of extra work when it's not necessary.

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The equations of motion give results along one direction and one had to enter into the equation all the vectors along that direction.When we resolve g along the direction of initial velocity "u" in the equation s = ut - 1/2gt^2 as gsine theta then we get the displacement s along the direction of u.Hence after the time the velocity of the object becomes zero along the direction of u the further use of the equation yield the reduced displacement along the direction u.

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