JEE Main 2015 · 4 Apr · Shift 1 · Physics Q1

← JEE Main 2015 · 4 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 1 of 90

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?

(Assume stones do not rebound after hitting the ground and neglect air resistance, take $g=10$ m/s $^2$ )

(The figures are schematic and not drawn to scale)

Rises linearly to 240 m at t = 8 s, then falls linearly to 0 at t = 12 s

Graph of (y₂ − y₁) in m against t(s): a straight rise to 240 at t = 8, then a straight fall to 0 at t = 12.

Rises linearly to 240 m at t = 12 s, then drops vertically to 0

Graph of (y₂ − y₁) in m against t(s): a straight rise to 240 at t = 12, then a vertical drop.

Rises linearly to 240 m at t = 8 s, then falls along a downward-curving arc to 0 at t = 12 s

Graph of (y₂ − y₁) in m against t(s): a straight rise to 240 at t = 8, then a curve bending downward (concave down) to 0 at t = 12.

Rises linearly to 240 m at t = 8 s, then falls along an upward-curving (concave) arc to 0 at t = 12 s

Graph of (y₂ − y₁) in m against t(s): a straight rise to 240 at t = 8, then a curve concave up falling to 0 at t = 12.

Answer and worked solution

Official answer: C

Take up as positive from the cliff top: $y_1=10t-5t^2$ and $y_2=40t-5t^2$ .

While both are in flight, $y_2-y_1=30t$ , a straight line.
Stone 1 lands ( $y_1=-240$ ) when $5t^2-10t-240=0$ , i.e. $t=8$ s, when $y_2-y_1=240$ m.
After that $y_1=-240$ , so $y_2-y_1=40t-5t^2+240$ . This is a downward-curving (concave-down) parabola.
It reaches 0 when stone 2 lands: $5t^2-40t-240=0$ , i.e. $t=12$ s.

That is graph (3).

Topics:Motion graphs Relative velocity

Next question →

JEE Main 2015 · 4 Apr · Shift 1, Physics question 1