Question:
A projectile is fired at an angle of $ 45^{\circ} $ and reaches the highest point in its path after $ 2\sqrt2\,s $ . Find the velocity of projectile in $ m s ^{-1} $ ,
A projectile is fired at an angle of $ 45^{\circ} $ and reaches the highest point in its path after $ 2\sqrt2\,s $ . Find the velocity of projectile in $ m s ^{-1} $ ,
Updated On: Jun 14, 2022
- $ 19.6 $
- $ 39.2 $
- $ 9.8 $
- $ 4.9 $
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The Correct Option is B
Solution and Explanation
Given : Time taken to reach highest point,
$t=2\sqrt{2}s$
Angle of projection, $\theta=45^{\circ}$
Let $u$ be the speed of projection
$\therefore t=\frac{u\,sin\,\theta}{g}$ or $u=\frac{gt}{sin \,\theta}$
or $u=\frac{2\sqrt{2}\times9.8}{\frac{1}{\sqrt{2}}}$
$=39.2\,m\,s^{-1}$
$t=2\sqrt{2}s$
Angle of projection, $\theta=45^{\circ}$
Let $u$ be the speed of projection
$\therefore t=\frac{u\,sin\,\theta}{g}$ or $u=\frac{gt}{sin \,\theta}$
or $u=\frac{2\sqrt{2}\times9.8}{\frac{1}{\sqrt{2}}}$
$=39.2\,m\,s^{-1}$
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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