Motion in One Dimension Problem - 4

Description:

Question

Two stones are thrown up simultaneously from the edge of cliff 200m height with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in figure, correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground take g = 10 ms². Give the equation for the linear and curved parts of the plot.

Solution

Position at any time is given by −

S = ut + 1/2 at²

For the first stone −

x₁ = 15t - 1/2 gt²

For the second stone −

x₂ = 30t - 1/2 gt²

Now,

(x₂ - x₁) = 15t

x₂ - x₁/t = +15

This indicates positive slope that results in a straight line portion OA.

First stone reach ground at time −

x₁ = 15t - 1/2 gt²

-200 = 15t - 1/2 × 10t²

-200 = 15t - 5t²

-40 = 3t - t²

t² - 3t - 40 = 0

t = 8 or -5

∴ t = 8 sec

After first stone reach the ground (v=0) and only second stone is moving.

(x₂ - x₁) = (30t - 1/2 gt²) - 0

This generates shape parabola

Second stone reach ground at time −

x₂ = 30t - 1/2 gt²

-200 = 30t - 1/2 × 10t²

-200 = 30t - 5t²

-40 = 6t - t²

t² - 6t - 40 = 0

t = 10 or -4

∴ t = 10 sec

When second stone reach the ground −

x₂ - x₁ = 0

Therefore the graph remain as it is.