Question:
A projectile is projected with speed $49\,ms^{-1}$ at an angle of $30^{\circ}$ with horizontal. The time taken by the projectile to reach the highest point is
A projectile is projected with speed $49\,ms^{-1}$ at an angle of $30^{\circ}$ with horizontal. The time taken by the projectile to reach the highest point is
Updated On: Jul 2, 2022
- $2\,s$
- $2.5\,s$
- $3.5\,s$
- $4.5\,s$
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The Correct Option is B
Solution and Explanation
Here,
Velocity of projection, $u=49\, m \, s^{-1}$
Angle of projection, $\theta=30^{\circ}$
Time taken by the projectile to reach the highest point is
$t=\frac{u\,sin\,\theta}{g}$
$=\frac{\left(49\,m\,s^{-1}\right)\left(sin\,30^{\circ}\right)}{9.8 \, m \,s^{-2}}$
$=2.5 \, s$
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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