mechanics problem (1 Viewer)

[cormglakes]

How do you do part b?
An object falls from the top of a building. An office worker sees the object fall past a window 2 m high in 1 5 s. Assume that air resistance is negligible and use g = 10 m/s^2 . Let y be the height of the object above the bottom of the window at time t.
(a) Show that y = H − 5t^2 , where H is the distance from the bottom of the window to the top of the building.
(b) Use part (a) and the information given in the question to form a pair of equations. Solve these simultaneously to find the value of H

Answer for b is 6.05m


Also another question :

so I got one of the answers which is T=mgcos(theta); but how do you get the other answer which is a=(-gsin(theta))/L

 

[CM_Tutor]

Also another question :

so I got one of the answers which is T=mgcos(theta); but how do you get the other answer which is a=(-gsin(theta))/L

If you continue from resolving the forces, you can add the accelerations in the x and y directions to get an overall acceleration of .

However, what you are seeking is , which follows as the arc length (call it ) is given by

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