Question:
An aeroplane flying horizontally with a speed of $360\,km\,h^{-1}$ releases a bomb at a height of $490\,m$ from the ground. If $g = 9.8\,m \,s^{-2}$, it will strike the ground at
An aeroplane flying horizontally with a speed of $360\,km\,h^{-1}$ releases a bomb at a height of $490\,m$ from the ground. If $g = 9.8\,m \,s^{-2}$, it will strike the ground at
Updated On: Jul 6, 2022
- $10 \,km$
- $100 \,km$
- $1 \,km$
- $16 \,km$
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The Correct Option is C
Solution and Explanation
Time taken by the bomb to fall through a height of $490\,m$
$t=\sqrt{\frac{2h}{g}}$
$=\sqrt{\frac{2\times490}{9.8}}=10\,s$
Distance at which the bomb strikes the ground
= horizontal velocity $\times$ time
$=360\,km\,h^{-1}\times10\,s$
$=360\,km\,h^{-1}\times \frac{10}{3600}h$
$=1\,km$
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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