Question

A ball is projected with a velocity 20 ms at an angle to the horizontal. In order to have the maximum range is velocity at the highest position must be close to



a.10ms



b.14 ms



d.16 ms



c.18 ms

Answer 1

Answer:

Hence the required answer is 14mm/s.

Answer 2

Answer:

The velocity at the highest position is 14.14m/s.

Explanation:

The range of a projectile is given as,

$R=\frac{u^2sin2\theta}{2g}$      (1)

Where,

R=range of the projectile

u=velocity with which the projectile is projected=20m/s (given)

θ=angle of projection

g=acceleration due to gravity=10m/s²

The range will be maximum when,

$sin2\theta=1$

$2\theta=90\textdegree$           (∵$sin90\textdegree=1$)

$\theta=\frac{90\textdegree}{2}=45\textdegree$      (2)

And the velocity of the projectile at the highest position will only be along the horizontal direction. i.e.

$v=ucos\theta$   (3)

By substituting the value of u and θ in equation (3) we get;

$v=20\times cos45\textdegree$

$v=20\times \frac{1}{\sqrt{2} }$             (∵$cos45\textdegree=\frac{1}{\sqrt{2} }$)

$v=14.14m/s$

Hence,  the velocity at the highest position is 14.14m/s.