Question:
A shell fired from a gun at sea level rises to a maximum height of $ 5\, km $ when fired at a ship $ 20 \,km $ away. The muzzle velocity should be
A shell fired from a gun at sea level rises to a maximum height of $ 5\, km $ when fired at a ship $ 20 \,km $ away. The muzzle velocity should be
Updated On: Jun 14, 2022
- $ 7 \,m/s $
- $ 14 \,m/s $
- $ 28 \,m/s $
- $ 56 \,m/s $
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The Correct Option is D
Solution and Explanation
$H = 5000\,m = \frac{u_y^2}{2g} $
$ = \frac{u_y^2}{2\times 10}$
$u_y^2 = 100000 $
or $u_y = 100 \sqrt{10} \frac{m}{s}\,\,\,...(i)$
$R = 20, 000\,m = \frac{2u_x \,u_y}{g}$
$ = \frac{2\times u_x}{10} \times 100\sqrt{10}$
or $u_x = 100\sqrt{10}\,\,\,...(ii)$
Now, $ u = \sqrt{u_x^2 + u_y^2} $
$= 100\sqrt{20}\,m/s$
$ = \frac{u_y^2}{2\times 10}$
$u_y^2 = 100000 $
or $u_y = 100 \sqrt{10} \frac{m}{s}\,\,\,...(i)$
$R = 20, 000\,m = \frac{2u_x \,u_y}{g}$
$ = \frac{2\times u_x}{10} \times 100\sqrt{10}$
or $u_x = 100\sqrt{10}\,\,\,...(ii)$
Now, $ u = \sqrt{u_x^2 + u_y^2} $
$= 100\sqrt{20}\,m/s$
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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