Question:
A hiker stands on the edge of a cliff $490\,m$ above the ground and throws a stone horizontally with a speed of $15\,ms^{-1}$. The time taken by the stone to reach the ground is
A hiker stands on the edge of a cliff $490\,m$ above the ground and throws a stone horizontally with a speed of $15\,ms^{-1}$. The time taken by the stone to reach the ground is
Updated On: Jul 5, 2022
- $5\,s$
- $10\,s$
- $12\,s$
- $15\,s$
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The Correct Option is B
Solution and Explanation
Here, $u = 15\,m \,s^{-1}$,
$h = 490\,m$
Time taken by the stone to reach the ground is
$t=\sqrt{\frac{2h}{g}}$
$=\sqrt{\frac{2\times490\,m}{9.8\,ms^{-2}}}$
$=10\,s$
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Notes on Motion in a plane
Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration