Question

A bullet fired at an angle of 60° with the vertical hits the ground at distance of 2km . Calculate the distance at which the bullet will hit the ground when fired at an angle of 45° , assuming the speed to be the same .​

Answer 1

we know , for a projectile motion

range = (u² sin2α)/g

where

α is the angle of projection with the horizontal

here α 30° ( given 60° is the angle made with vertical so 30° is the angle made with horizontal)

2000 m = ( u² sin2×30°)/ g. g=10 m/s²

u²= (2000 ×10×√3 ) /2. sin 60° =√3/2

u²=1000 ×10×√3

u² = 10000√3

now if α=45°

range = (u² sin2×45)/g

substituting the value of u² in above equation

range = (10000√3×1)/10

range = 1000√3

may be this can be your answer

hope it helps you

Answer 2

Given: A bullet fired at an angle 60° with the vertical hits the ground at a distance of 2 km.

To find: Range of projectile assuming speed to be same and angle 45°

Explanation: The formula for calculating range of projectile is given by the formula:

$range = \frac{{u}^{2} sin(2 \alpha )}{g}$

where α is angle made with the vertical and g is acceleration due to gravity.

In first case,α = 60° then 2α= 120°

sin 120° = 1/2

= 0.5

Range= u^2 * 0.5 / 10

=> 2 km = u^2 / 20

=> u^2 = 40 km/s

In second case, α= 45° then 2α= 90°

sin 90° = 1

Range in this case= u^2 * 1/10

= 40/10

= 4 km

Therefore, the distance at which the bullet will hit the ground when angle made with the vertical is 45° is 4 km.