Question:
A ball is projected horizontally with a velocity of $5 \,m/s$ from the top of a building $19.6\, m$ high. How long will the ball take of hit the ground ?
A ball is projected horizontally with a velocity of $5 \,m/s$ from the top of a building $19.6\, m$ high. How long will the ball take of hit the ground ?
Updated On: Jan 12, 2024
- $\sqrt{2}\,s$
- $2\, s$
- $\sqrt{3}\,s$
- $3\, s$
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The Correct Option is B
Solution and Explanation
The time taken to hit the ground is given by
$T=\sqrt{\frac{2 H}{g}}$
$=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \,s$
$T=\sqrt{\frac{2 H}{g}}$
$=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \,s$
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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