The problem asks for the time(s) when a ball, thrown vertically upwards from a moving car, reaches a specific height relative to the car.

Irrelevant Information Handling

The car moves horizontally at a constant velocity of 3 m/s. This horizontal motion does not affect the vertical motion of the ball relative to the car. Therefore, the car's velocity is irrelevant for this calculation.

Vertical Motion Kinematics

We focus solely on the ball's vertical motion. The relevant kinematic equation relating vertical displacement ($y$), initial vertical velocity ($v_0$), acceleration ($a$), and time ($t$) is:

$y = v_0 t + \frac{1}{2} a t^2$

Given values are:

• Initial vertical velocity, $v_0 = 20$ m/s (upwards)
• Acceleration due to gravity, $a = -10$ m/s$^2$ (downwards)
• Target vertical displacement (height relative to the car's top), $y = 15$ m

Calculating Time

Substitute the values into the kinematic equation:

$15 = (20) t + \frac{1}{2} (-10) t^2$

Simplify the equation:

$15 = 20t - 5t^2$

Rearrange the equation into the standard quadratic form ($At^2 + Bt + C = 0$):

$5t^2 - 20t + 15 = 0$

Divide the entire equation by 5 to simplify:

$t^2 - 4t + 3 = 0$

Factor the quadratic equation:

$ (t - 1)(t - 3) = 0 $

Solving for $t$, we get two possible values:

$ t - 1 = 0 \implies t = 1 \text{ s} $

$ t - 3 = 0 \implies t = 3 \text{ s} $

Conclusion

The ball is at a height of 15 m from the top of the moving car at times $t = 1$ second and $t = 3$ seconds. These correspond to options A and C.