Question

ram throws a small water balloon from ground so as to hit shyam shyam standing on top of a building height 10 m if the building is at distance 20 m away from ram and throws the balloon at angle 45 with the horizontal the speed of projection is

Answer 1

Based on the given information and using the equation for range in projectile motion, the initial speed at which Ram throws the water balloon is approximately 14.5 m/s.

Step-by-step explanation:

To find the initial speed at which Ram throws the water balloon, we need to analyze the projectile motion.

The horizontal distance traveled by the balloon can be found using the formula:

Range (R) = (Initial Speed (v)² * sin(2θ)) / g

In this case, the range (R) is equal to the distance between the building and Ram, which is 20 m.

The angle (θ) is 45°. The acceleration due to gravity (g) is approximately 9.8 m/s².

Substituting these values into the formula, we have:

20 = (v² * sin(2 * 45°)) / 9.8

Simplifying the equation, we get:

20 = (v² * sin(90°)) / 9.8

Sine of 90° is equal to 1, so the equation becomes:

20 = v² / 9.8

Multiplying both sides of the equation by 9.8, we get:

v² = 20 * 9.8

Finally, taking the square root of both sides of the equation, we find:

v ≈ √(20 * 9.8)

≈ 14.5 m/s