# mechanics problem (1 Viewer)

#### cormglakes

##### New Member
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

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

##### Moderator
Moderator
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 $\bg_white -g\sin\theta$.

However, what you are seeking is $\bg_white \ddot{\theta} = -\cfrac{g}{L}\sin\theta$, which follows as the arc length (call it $\bg_white r$) is given by

$\bg_white r = L\theta \quad \implies \quad \dot{r} = L\dot{\theta} \quad \implies \quad \ddot{r} = L\ddot{\theta} \quad \implies \quad \ddot{\theta} = \cfrac{\ddot{r}}{L} = -\cfrac{g\sin\theta}{L}$