Question

A ball is projected from ground with some speed at an angle from horizontal. The power of force of gravity for the ball during journey is

Answer 1

Answer:

A body of man 'm' projected with speed U at angle θ

with horizontal

So maximum height H=  

2g

u  

2

sin  

2

θ

​  

 

half of maximum height  

2

H

​  

=  

4g

u  

2

sin  

2

θ

​  

 

vertical component of initial velocity u  

y

​  

=usinθ

v  

y

2

​  

=u  

y

2

​  

+2a  

y

​  

y

⇒v  

y

2

​  

=(usinθ)  

2

−29(  

49

u  

2

sin  

2

θ

​  

)

v  

y

2

​  

=  

2

u  

2

sin  

2

θ

​  

=v  

y

​  

=  

2

​  

 

usinθ

​  

 

Power = F  

y

​  

V  

y

​  

 

F  

y

​  

=mg

=  

2

​  

 

mgusinθ

​

Explanation:

Answer 2

Answer:

As we have derived the equation, the power of force of gravity of the ball during the journey is $F = mgcos \theta$

Explanation:

Given - a ball is projected from the ground
           - a specific angle (θ) of projection from the horizontal

To find - the force of gravity for the ball

Solution - Let us consider the mass of the ball to be 'm'.

                 The maximum height the ball reaches from the horizontal would be

So maximum height $H = \frac{2u sin \theta}{g}$

Thus, half of the maximum height would be $H_\frac{1}{2} = \frac{u sin \theta}{g}$

Next, we can write the vertical part of the initial velocity to be $u_y = u sin \theta$ and the horizontal part of the same to be $u_x = u cos \theta$.

Thus, we can write the initial velocity to be $u = mg$, where g is the acceleration due to gravity.

Now, since the force of gravity acts only along the vertical direction, we can write it as $F = mg cos \theta$

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