Projectile Motion

The maximum height attained by a projectile can be derived from the vertical motion component of the projectile. The vertical velocity at the maximum height is zero, and the equation of motion is: where: is the vertical component of the initial velocity, at the maximum height, is the acceleration due to gravity, is the maximum height. Rearranging for , we get: Let the two projectiles have angles of projection: For the first projectile: For the second projectile: The ratio of their maximum heights is: Canceling common terms and substituting and , we get: Simplify: Thus, the ratio of the maximum heights is: Final Answer:

Step 1: Horizontal range of a projectile
The horizontal range of a projectile is given by: where is the initial speed, is the angle of projection, and is the acceleration due to gravity. Step 2: Maximum horizontal range
The maximum horizontal range occurs when , i.e., . Then: Step 3: Area of spread
Assuming bullets are fired in all directions, the spread forms a circle with radius . The area is:

The increase in volume is given by: where: (initial volume of the cube), (coefficient of volumetric expansion), (temperature change). The surface area of a cube is: Given , solve for : The initial volume is: Substitute into the formula for : Simplify:

The formula for the acceleration due to gravity at a height above the Earth's surface is: Given: Substituting into the equation: Simplifying: Taking the square root: Solving for : Substituting : Simplifying further: Final Answer:

We can use the first equation of motion to calculate the time taken to reach the maximum height: Where: - (final velocity at maximum height), - (initial velocity), - (acceleration due to gravity, acting downward), - is the time taken to reach maximum height. At maximum height, , so the equation becomes: Solving for : Thus, the time taken for the ball to reach its maximum height is approximately .

The object is dropped horizontally from the helicopter. In this case, the horizontal velocity of the helicopter will be the horizontal velocity of the object upon release. The vertical velocity can be found using the equation for free fall. The horizontal displacement is given by: Where is the horizontal velocity and is the time. From the problem, we know the horizontal velocity , and time . Converting to m/s: Thus, the horizontal displacement is: So the displacement relative to the helicopter is .

Concept:
In projectile motion, the maximum height and horizontal range depend on the initial velocity and the angle of projection. When the angle of projection is , the projectile achieves maximum horizontal range.

Step 1: Formula for maximum height For :

Step 2: Formula for horizontal range For :

Step 3: Find relation between and Conclusion:
Thus, when a projectile is fired at , the relation between range and maximum height is .