Question

A bullet is fired from a cannon with velocity 500m/s. If the angle of projection is 15 and g=10 then the range is

Answer 1

The range is 12.5 km.

Explanation:
Given: u=500 m/s
Theta=15
G=10

We know,

Range= (u x u x sin2theta)/ g

This will equal to 12500m= 12.5 km

Answer 2

Answer:

Range, R = 12500 meters

Explanation:

Given that,

Velocity of the bullet, v = 500 m/s

Angle of projection, $\theta=15$

Acceleration due to gravity, $g=10\ m/s^2$

We need to find the range of the projectile. its formula is given by :

$R=\dfrac{v^2\ sin2\theta}{g}$

$R=\dfrac{(500)^2\ sin(30)}{10}$

R = 12500 meters

So, the range of the projectile is 12500 meters. Hence, this is the required solution.