Question

Help solve pls?!

Answer 1

At the point A initially the object was in rest. So its initial velocity `u=0`,
The component of acceleration due to gravity acting downward along the inclined plane is `gsin30^@=10*1/2=5m"/"s^2`

Again the distance

`AB =("height of A from B"/sin30^@ )m=1/(1/2)=2m`

Hence velocity gained by the object at B will be

`v=(sqrt(u^2+2*gsin30^@*2)) m"/"s`

`=>v=sqrt(20) m"/"s`

At B the horizontal component of velocity `v_x=vcos30^@=sqrt20*sqrt3/2=sqrt15m"/"s`

And the downward vertical component will be

`v_y=vsin30^@=sqrt20*1/2=sqrt5m"/"s`

If time taken by the object to reach at P from B be `t` then considering the vertical displacement of the object from B to P we can write

`10=v_y*t+1/2*g*t^2`

`=>10=sqrt5t+1/2*t^2`

`=>t^2+2sqrt5t-20=0`

`=>t=(-2sqrt5+sqrt((2sqrt5)^2+4*20))/2`

`=>t=(5-sqrt5)s`

So the horizontal displacement during time of flight `t` will be

`X=v_x*t=sqrt15(5-sqrt5)m`

Answer 2

See below.

Explanation:

The movement is mechanically conservative then at point A the mechanical energy regarding the floor level is

`A -> E = h_A m g` Here `h_A = 11` [m]

At the release point is

`B->E = h_B m g + 1/2m v_B^2` Here `h_B = 10`

then

`h_A m g = h_B m g + 1/2m v_B^2` and

`v_B = sqrt(2(h_A-h_B)g) = 2 sqrt(5) = 4.47` [m/s] --- (i)

After the release, considering the coordinate system origin at the release point we have

`(x,y) = ((v_B cos theta)t, -(v_B sin theta)t-1/2g t^2)` and the floor is reached at `t_f` such that

`y = -(v_B sin theta)t_f-1/2g t_f^2 = -h_B` or

`t_f=(-sintheta v_B + sqrt[2 g h_B + sintheta^2 v_B^2])/g = 1/2(sqrt(205)-sqrt(5)) = 1.21` [s] --- (ii)

and the floor projected flight distance is

`X = (v_B cos theta)t_f= 1/2 sqrt[3/5] ( sqrt[205]-sqrt[5] ) = 4.68` [m] --- (iii)