Answer:
A. The range of A and B are equal.
Explanation:
Let u be the initial speed of both the projectile.
The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.
For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,
The speed in the horizontal direction
and the speed in the vertical direction is upward.
For A:
The speed in the horizontal direction
and the speed in the vertical direction is upward.
For B:
The speed in the horizontal direction
and the speed in the vertical direction is upward.
Let and are the time of flight for projectile A and B respectively.
As the range is the horizontal distance traveled by the projectile, so
The range for the projectile A
The range for the projectile B =
At the highest point, the vertical velocity is 0.
Bu using the equation of motion
Here, the final velocity v=0, the initial velocity , h= vertical distance up to the highest point, and (as per sign convention).
So,
For projectile A: The maximum height attained.
For projectile B: The maximum height attained.
As , the height of A is attained by A is more than the heigHt attained by B.
Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.
So, the total time of flight = 2 x (Time to reach the highest point)
In a similar way, by using the equation of motion v=u+at,
The time to reach the highest point
where g is the acceleration due to gravity.
So, the total time of flight
The total time of flight for A
The total time of flight for A
Now, from equations (i) and (ii),
The range for the projectile A =
The range for the projectile B =
Both the projectile have the same range.
Hence, option (A) is correct.