About fall motio

Question

from a building 40m above the ground an object is released without initial velocity. at a height of 25 m the object gets a new velocity of 20 m / s with the direction of theta = 60. calculate the magnitude and direction of the velocity of the object when it reaches the ground

Answer 1

I have another interpretation of this problem. The object is allowed to fall with no initial downward velocity. At a point 15 m from the top of the building, it would have a downward velocity of 17.1 m/s. At that point, another velocity is imparted to the object, perhaps from a gust of wind. This new velocity has magnitude 20 m/s in a direction of 60º from the vertical. This new velocity has a downward component equal to 20cos 60 (10.0 m/s), and a horizontal component 20sin60 (17.3 m/s). The new vertical component would add to the 17.1 m/s that the object has due to the fall. The final velocity at the ground would have a vertical component obtained from the equation

Vv² = Vo² + 2x9.8x25 where V_o is the vertical component of the combined gravitationally caused velocity and the vertical component of the new velocity ( combined 27.1 m/s). I calculated the final vertical velocity component to be 35.0 m/s. The horizontal component at the ground is 17.3 m/s. This leads to a velocity vector of magnitude 39.0 m/s at an angle of 26.3º from the vertical.

Please comment if or when you know the actual answer. Thanks.

Answer 2

from a building 40m above the ground an object is released without initial velocity. at a height of 25 m the object gets a new velocity of 20 m / s with the direction of theta = 60. calculate the magnitude and direction of the velocity of the object when it reaches the ground

Here initial velocity is zero

Initial height = 40m

At a height of 25 m , it has fallen through 15 m

Its vertical component of velocity at this pint = V cos theta = V Cos 60 =20 Cos 60 =10m/s

So V = 10 m/s for a fall of 15 m

using V² = 2aS

Gives a= V²/2S = 10*10/ 2*15 = 100/30 = 3.33m/s

Final vertical velocity when touching the ground = Sqrt ( 40 +2*3.33*25)= sqrt (206.5) =14.37 m/s =Vy

But it has one more component as horizontal = V sin theta = VSin 60 = 10*0.577 = 5.77 m/s = Vx

Now find the resultant of Vx and Vy

resultan = sqrt ( Vx²+vy² )

tan theta = Vy/Vx Gives direction