Two stoves are thrown up simultaneously from the edge of a child 200 m

1.	Two stoves are thrown up simultaneously from the edge of a child 200 m high with initial speeds of 15 ms-1 and 30 ms-1. verify that the graph shown in the figure correctly represents the time variation of the relative position of the second stone w.r.t. the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms-2. Give the equation for the linear and curved parts of the plot.
Answer» For first stone, x(0) = 200 m, v(0) = 15 ms^-1 a = -10 ms-2 x₁(t) = x(0) + v(0)t +1/2 at² x₁(t) = 200 + 15t - 5t² When the first stone hits the ground, x1 (t) = 0 or, -5t² +15 t² + 200 = 0 On simplification, we get t = 8s. For second stone, x(0) = 200 m, v(0) = 30 ms^-1, a= -10 ms^-2 x₂(t) = 200 +30t -5t² When this stone hits the ground x₂ (t) = 0 or, -5t² + 30t +200 = 0 Relative position of second stone w.r.t, first is given by x₂(t) -x₁(t) = 15t. Since, there is a linear relationship between x₂(t) x₁ (t) and t, the graph is a straight line. For maximum separation, t = 8t so. maximum separation is 120 m. After 8 second, only the second stone would be in motion. so, the graph is in accordance with the quadratic equation.

Two stoves are thrown up simultaneously from the edge of a child 200 m high with initial speeds of 15 ms-1 and 30 ms-1. verify that the graph shown in the figure correctly represents the time variation of the relative position of the second stone w.r.t. the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms-2. Give the equation for the linear and curved parts of the plot.

Answer»

For first stone,

x(0) = 200 m, v(0) = 15 ms^-1

a = -10 ms-2

x₁(t) = x(0) + v(0)t +1/2 at²

x₁(t) = 200 + 15t - 5t²

When the first stone hits the ground, x1 (t) = 0

or, -5t² +15 t² + 200 = 0

On simplification, we get t = 8s.

For second stone, x(0) = 200 m, v(0) = 30 ms^-1,

a= -10 ms^-2

x₂(t) = 200 +30t -5t²

When this stone hits the ground x₂ (t) = 0

or, -5t² + 30t +200 = 0

Relative position of second stone w.r.t, first is given by

x₂(t) -x₁(t) = 15t.

Since, there is a linear relationship between x₂(t) x₁ (t) and t, the graph is a straight line.

For maximum separation, t = 8t

so. maximum separation is 120 m.

After 8 second, only the second stone would be in motion. so, the graph is in accordance with the quadratic equation.