Two stones A and B projected with speed u and square root of 2u attain...
Explanation:
Understanding the problem:
Two stones A and B are thrown with different speeds u and square root of 2u respectively. They both attain the same maximum heights. Stone A is thrown at an angle 60 degrees with horizontal. We need to find the approximate angle of projection of stone B with horizontal.
Understanding the concept:
The maximum height attained by a projectile depends on the angle of projection and speed of projection. If we throw a projectile at two different angles but with the same initial speed, the maximum height attained by the projectile will be different for the two cases. However, if we throw two projectiles with different initial speeds but they attain the same maximum height, then the angles of projection will be different.
Solution:
Let's assume that the maximum height attained by both the stones is h.
For stone A:
- Initial speed = u
- Angle of projection = 60 degrees
For stone B:
- Initial speed = square root of 2u
- Angle of projection = ? (let's assume it to be theta)
We know that the time taken by a projectile to reach maximum height is given by:
time = (u sin theta)/g
where g is the acceleration due to gravity.
Using this, we can calculate the time taken by both the stones to reach maximum height:
For stone A:
time = (u sin 60)/g = (u/2g) seconds
For stone B:
time = [(square root of 2u) sin theta]/g = [(square root of 2) sin theta]/g seconds
Now, we can use the formula for maximum height to equate the maximum heights of both the stones:
Maximum height = (u^2 sin^2 theta)/(2g) = ((square root of 2u)^2 sin^2 theta)/(2g) = h
Simplifying this equation, we get:
sin^2 theta = (2/3)
Taking the square root, we get:
sin theta = square root of (2/3)
Using the inverse sine function, we get:
theta = 37 degrees (approx)
Answer:
The approximate angle of projection of stone B with horizontal is 37 degrees.