Gravity is the only one helping it.
The ball's horizontal position in the air is

It hits the wall when
, which happens at

Meanwhile, the ball's vertical position is

where
is the acceleration due to gravity, 9.80 m/s^2.
At the time the ball hits the wall, its vertical position (relative to its initial position) is

Answer:
0.3625
Explanation:
From the given information:
Consider the equilibrium conditions;
On the ladder, net torque= 0
Thus,
; and

However, by rearrangement;

![\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}](https://tex.z-dn.net/?f=%5Cmu%3D%20%5Cdfrac%7B%20%28323%20%20%5C%20N%29%20%28%2010.8%20%5C%20m%29%20%5C%20cos%2056%5E0%20%2B%20%28734%20%5C%20N%29%20%287.46%20%5C%20m%29%20%5C%20cos%20%20%5C%20%2066.46%5E0%7D%7B%5CBig%20%5B%28323%20%5C%20N%29%2B%28734%20%5C%20N%29%20%5CBig%5D%20%2810.8%20%5C%20m%29%7D)

Answer:
Explanation:
As the field of current wire passing through the loop is same in direction (normally inward) but not uniform in magnitude.
So, we will use integtation method for finding the flux.
The same flux through a thin reactangular strip of length l and with dx, is given by
dϕB=Bx.dS=B(x)dScos180∘.
Magnetic field due to a long straight wire carrying current I is given by B=μ0I2πx, and area, dS=l×dx.
∴ ϕB=∫dϕB=−∫μ02πIlx.dx=−μ02πIl[logex]x=a+lx=a
=−μ02π.Ill