Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
Answer:
Option C
100 J
Explanation:
Kinetic energy, KE is given by
where m is the mass and v is the velocity
Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain
Therefore, the child's kinetic energy is equivalent to 100 J
Newton's 2nd law of motion:
Net Force = (mass) x (acceleration) .
The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.
If you know any two of the quantities in the formula,
the law can be used to calculate the third one.
Answer: Place his feet parallel to the baseline prior to tossing the ball
