Question

A particle is dropped from top of tower of 40m and simultaneously another particle is projected from ground in upward direction with 20 m/s! Find when and at what distance from ground they will meet C

Answer 1

They will meet at 2 seconds and at a distance of 20 m from the ground.

Answer 2

Let the origin be at the ground and the upward direction be positive. The height of the tower is $H = 40$ m. We assume the acceleration due to gravity is $g = 10 \, m/s^2$.

For the particle dropped from the top of the tower:

• Initial position $y_1(0) = H = 40$ m.
• Initial velocity $v_1(0) = 0$ m/s (since it is dropped).
• Acceleration $a_1 = -g = -10 \, m/s^2$ (since gravity acts downwards).

The position of the particle at time $t$ is given by the equation of motion: $y_1(t) = y_1(0) + v_1(0)t + \frac{1}{2}a_1t^2$ $y_1(t) = 40 + 0 \cdot t + \frac{1}{2}(-10)t^2$ $y_1(t) = 40 - 5t^2$

For the particle projected from the ground:

• Initial position $y_2(0) = 0$ m.
• Initial velocity $v_2(0) = 20$ m/s (upwards).
• Acceleration $a_2 = -g = -10 \, m/s^2$ (since gravity acts downwards).

The position of the particle at time $t$ is given by the equation of motion: $y_2(t) = y_2(0) + v_2(0)t + \frac{1}{2}a_2t^2$ $y_2(t) = 0 + 20t + \frac{1}{2}(-10)t^2$ $y_2(t) = 20t - 5t^2$

The two particles meet when their positions are the same, i.e., $y_1(t) = y_2(t)$. $40 - 5t^2 = 20t - 5t^2$ $40 = 20t$ $t = \frac{40}{20}$ $t = 2$ seconds.

So, the particles meet after 2 seconds.

To find the distance from the ground where they meet, we can substitute $t = 2$ seconds into either $y_1(t)$ or $y_2(t)$. Using $y_1(t)$: $y_1(2) = 40 - 5(2)^2 = 40 - 5(4) = 40 - 20 = 20$ meters. Using $y_2(t)$: $y_2(2) = 20(2) - 5(2)^2 = 40 - 5(4) = 40 - 20 = 20$ meters.

Both equations give the same meeting point, which is 20 meters from the ground.