Question

9. A bullet is fired from the gun, it covers maximum horizontal distance of 10 km. Determine the
velocity of projection and the maximum height attained by the bullet.
Ans. 313 m/s, 2492 m​

Answer 1

Answer:

313 m/s

2500 m

Explanation:

A bullet is fired from the gun, it covers maximum horizontal distance of 10 km. Determine the

velocity of projection and the maximum height attained by the bullet.

To  cover maximum horizontal distance angle of Projection should be 45 deg

Let Say Velocity was V√2  m/s

Horizontal Velocity =  V√2 Cos45 = V m/s

Vertical Velocity = V√2 Sin45 = V m/s

Using V = U + at

a = g = 9.8 Vertical Velocity at top = 0

Time to reach max height = V/g = V/9.8  sec

Time of Flight = 2V/9.8 = V/4.9

Horizontal Distance = V * V/4.9  = 10000

=> V = 100√4.9

=> V√2 = 100√9.8 = 313 m/s

using V² - U² = 2aS

Height = V²/2g   = 49000/(2 * 9.8) = 2500 m

Answer 2

The maximum horizontal range :

The Horizontal distance covered by the projectile when it is projected at an angle 45 degrees. Because, Range = u²sin2θ/g. sin2θ has a maximum value when theta is 45 degrees.

Firing of a bullet is a projectile.

Given, Horizontal range of the bullet = 10km.

From the relation,

$\large {R = \frac {u^2 sin2 \theta} {g}}$

We have, sin2θ = 1, g = 9.8 m/s²

$R = 10 \times {10}^{3} = \frac{ {u}^{2} \times 1}{g} \\ \\ \implies \: 10000 \times 9.8 = {u}^{2} \\ \\ \implies \: u = \sqrt{98000} \\ \\ \implies \: u \: = 313 \: m {s}^{ - 1}$

Maximum Height :

The maximum vertical distance reached a projectile when it's velocity becomes zero.

$H_{max} = \frac{u ^{2} sin ^{2} \theta} {2g} \:$

$H_{max} = \frac{98000 \times (\frac{1}{ \sqrt{2}}) ^{2} }{2 \times 9.8} \\ \\ H_{max} = \frac{10000}{4} = 2500m ≈ 2492 m$

The velocity of projection and the maximum height attained by the bullet are 313 m/s, 2492 m