To analyze the situation where a boy throws a ball at an angle of 45 degrees and then starts running along the x-axis, we need to break down the motion of both the ball and the boy. The trajectory of the ball is given by the equation \( y = x - 5x^2 \), and the boy runs with a constant velocity of 1 m/s. Let’s evaluate each of the options provided to determine which are correct.
Understanding the Motion of the Ball
The equation of the ball's trajectory, \( y = x - 5x^2 \), describes a parabolic path. Here, the horizontal position \( x \) and vertical position \( y \) of the ball are related. The term \( -5x^2 \) indicates that the ball will eventually fall back down after reaching a certain height.
Calculating the Boy's Position
The boy starts at the origin (0, 0) and moves along the x-axis with a constant velocity of 1 m/s. Therefore, at any time \( t \), his position can be expressed as:
- Boy's position: \( (x_b, y_b) = (t, 0) \)
Analyzing the Options
Now, let’s evaluate each option one by one:
Option (a): Boy and ball are in the same vertical line at each instant of time.
For the boy and the ball to be in the same vertical line, their x-coordinates must be equal. The ball's x-coordinate is \( x \), and the boy's x-coordinate is \( t \). Since the ball's x-coordinate is not necessarily equal to \( t \) at all times (it depends on the trajectory), this statement is false.
Option (b): At t = 0.1 s, the velocity of the ball w.r.t. boy = zero.
To find the velocity of the ball, we need to differentiate the trajectory equation with respect to time. The horizontal and vertical components of the ball's velocity can be derived from its trajectory. The horizontal velocity \( v_x \) is constant (since there’s no horizontal acceleration), and the vertical velocity \( v_y \) can be found using the derivative of \( y \) with respect to \( x \) and the chain rule. At \( t = 0.1 \) s, we can calculate the velocities and then find the relative velocity with respect to the boy. This option requires detailed calculations but is likely false as the velocities will not be equal.
Option (c): At t = 0.2 s, the ball hits the boy.
To check if the ball hits the boy at \( t = 0.2 \) s, we need to find the position of both the ball and the boy at that time. The boy's position will be \( (0.2, 0) \). We can substitute \( x = 0.2 \) into the trajectory equation to find the ball's position. If the y-coordinate of the ball at \( x = 0.2 \) is 0, then they meet. Calculating this gives:
- For \( x = 0.2 \): \( y = 0.2 - 5(0.2)^2 = 0.2 - 0.2 = 0 \)
Since both the boy and the ball are at (0.2, 0), this option is true.
Option (d): Ball never hits the boy.
This option contradicts option (c). Since we found that the ball does hit the boy at \( t = 0.2 \) s, this statement is false.
Summary of Findings
Based on our analysis:
- Option (a) is false.
- Option (b) is likely false.
- Option (c) is true.
- Option (d) is false.
Thus, the only correct option is (c): At t = 0.2 s, the ball hits the boy.
