An elevator, in which a man is standing, is moving upward with a constant acceleration of 2 m/s2 . At some instant when speed of elevator is 10 m/s, the man drops a coin from a height of 1.5 m. Find the time taken by the coin to reach the floor.

Question

An elevator, in which a man is standing, is moving upward with a constant acceleration of 2 m/s2 . At some instant when speed of elevator is 10 m/s, the man drops a coin from a height of 1.5 m. Find the time taken by the coin to reach the floor. (A) (1/√3)s (B) (1/2)s (C) (1/√2)s (D) 1 s

Tardigrade Admin · Accepted Answer

drel = (1/2) arel t2 ⇒ 1.5 = (1/2)(10 + 2)t2 1.5 = (1/2) × 12 × t2 ⇒ t = (1/2) s

Q. An elevator, in which a man is standing, is moving upward with a constant acceleration of $2 m / s^{2}$ . At some instant when speed of elevator is $10 m / s$ , the man drops a coin from a height of $1.5 m$ . Find the time taken by the coin to reach the floor.

$\frac{1}{3} s$

$\frac{1}{2} s$

$\frac{1}{2} s$

1 s

$d_{re l} = \frac{1}{2} a_{re l} t^{2}$

$\Rightarrow 1.5 = \frac{1}{2} (10 + 2) t^{2}$

$1.5 = \frac{1}{2} \times 12 \times t^{2}$

$\Rightarrow t = \frac{1}{2} s$

Q. An elevator, in which a man is standing, is moving upward with a constant acceleration of 2m/s2 . At some instant when speed of elevator is 10m/s, the man drops a coin from a height of 1.5m. Find the time taken by the coin to reach the floor.

3​1​s

21​s

2​1​s

1 s

drel​=21​arel​t2 ⇒1.5=21​(10+2)t2 1.5=21​×12×t2 ⇒t=21​s

Q. An elevator, in which a man is standing, is moving upward with a constant acceleration of $2 m / s^{2}$ . At some instant when speed of elevator is $10 m / s$ , the man drops a coin from a height of $1.5 m$ . Find the time taken by the coin to reach the floor.

$\frac{1}{3} s$

$\frac{1}{2} s$

$\frac{1}{2} s$

$d_{re l} = \frac{1}{2} a_{re l} t^{2}$

$\Rightarrow 1.5 = \frac{1}{2} (10 + 2) t^{2}$

$1.5 = \frac{1}{2} \times 12 \times t^{2}$

$\Rightarrow t = \frac{1}{2} s$