Projectile Motion

Given the equation of the projectile's path: . Comparing this with the standard equation of a trajectory , we have:
(A)
(B)
(a) Range: The range is the value of when . or So, the range . Thus, a-i.
(b) Maximum height: The maximum height occurs at . Substituting this into the equation for : Thus, b-iii.
(c) Time of flight: From , we have . The time of flight . From , we have and . Thus, c-iv.
(d) Tangent of projection: From , the tangent of projection is . Thus, d-ii. Therefore, the correct matching is: a-i, b-iii, c-iv, d-ii.

Step 1: We are given: - Maximum range of the projectile = 80 m - Time of flight is unknown - Acceleration due to gravity, We need to find the distance traveled in the first 3 seconds.

Step 1: Maximum Range Formula
The maximum range of a projectile is given by the formula: For maximum range, and . Thus, Given , we can substitute the known values:

Step 2: Time of Flight
The time of flight is given by: Since and ,

Step 3: Horizontal Distance in 3 Seconds
The horizontal distance at time is given by: Since , Since the projectile follows a symmetric path, the distance covered in the first 3 seconds must be proportional to the total range. By symmetry,

Step 4: Final Answer
Final Answer: (2) 60 m

To determine the height from which the bowling machine releases the balls, we analyze the projectile motion and the given conditions. Given:
- Initial velocity
- All landing velocities make angles of 30° or more with the horizontal.

Step 1: Determine the Landing Velocity Components When a projectile lands, its velocity can be broken into horizontal ( ) and vertical ( ) components. The angle that the landing velocity makes with the horizontal is given by: Given that , we have:

Step 2: Express and in Terms of Initial Conditions The horizontal component of velocity remains constant: The vertical component of velocity at landing is: where is the launch angle and is the time of flight.

Step 3: Time of Flight The time of flight for a projectile launched from height is given by: Solving for involves solving a quadratic equation, but we can use the fact that the minimum angle is 30° to find the minimum height.

Step 4: Minimum Height Calculation For the minimum angle : Using the relationship between and : Since and , we can equate: Solving for and substituting back into the height equation, we find: For the minimum angle : However, this does not match the given options. Let's consider the maximum height for the given conditions.

Step 5: Re-evaluate the Approach Given the complexity, let's use the fact that the minimum height corresponds to the minimum angle. For : With : This still does not match the options. Let's consider the maximum height for the given conditions.

Step 6: Correct Calculation Given the discrepancy, let's use the energy approach. The total energy at launch equals the total energy at landing: Given and : Solving for :

This is closest to option (4). Final Answer:

Step 1: Understanding Maximum Height Condition For a projectile, the maximum height attained is given by: where is the initial velocity, is the angle of projection, and is the acceleration due to gravity.

Step 2: Maximum Range Projection For maximum range, the projectile is launched at , so the height attained is:

Step 3: Maximum Height Projection For maximum height, the projectile is launched vertically ( ), so the height attained is: Since both objects attain the same height, Solving for :