Question

No
A ball is launched by a man standing on the top of a building who holds the
ball at a distance 1m above the edge A with u=6m/s, along the horizontal
such that the ball just clears the edge C. Also find x , in meter, where the
Vi
3.9
C
100
A
-2
ball strikes the ground. Take g = 9.8ms-2​

Answer 1

Answer:

Hey Cute!

Explanation:

Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Now,  

(a). The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

 s=ut−  

2

1

​  

gt  

2

 

gt=2sinθ

t=  

g

2sinθ

​  

 

(b). The  horizontal range is

Horizontal range OA = horizontal component of velocity × total flight time

 R=ucosθ×  

g

2usinθ

​  

 

R=  

g

u  

2

sin2θ

​  

 

(c). The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

 v  

2

=u  

2

−2as

0=u  

2

sin  

2

θ−2gH

H=  

2g

u  

2

sin  

2

θ

​