On doubling the initial velocity of a projectile, its maximum hor
| On doubling the initial velocity of a projectile, its maximum horizontal range would now become
A. Two times greater
B. Four times greater
C. Four times reduced
D. Two times reduced
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
CONCEPT:
- Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
- Initial Velocity: The initial velocity can be given as x components and y components.
ux = u cosθ
uy = u sinθ
Where u stands for initial velocity magnitude and θ refers to projectile angle.
- Maximum Height: The maximum height is reached when vy = 0.
\({\rm{h}} = \frac{{{{\rm{u}}^2}{{\sin }^2}{\rm{\theta }}}}{{2{\rm{g}}}}\;\)
Where h is the maximum height.
- Range: The range of the motion is fixed by the condition y = 0.
\(R = \frac{{{u^2}sin2\theta }}{g}\)
Where R is the total distance covered by the projectile.
CALCULATION:
The range is given by:
\(R = \frac{{{u^2}sin2\theta }}{g}\)
When initial velocity (u) is doubled (2u):
The range will be:
\(R = \frac{{{(2u)^2}sin2\theta }}{g}\)
So new range (R') = 4R
Hence option 2 is correct.
