A bullet is fired with an initial velocity 300m s at an angle 30 degree with horizontal . At what distance from the gun will the bullet strike the ground ?
d = Vo^2*sin(2A)/g = 300^2*sin(60)/9.8 =
175m/s
R=u^2sin2(A)/g
R=150^2*3*sin60/9.8
To find the distance from the gun where the bullet will strike the ground, we can use the equations of motion.
First, let's break down the initial velocity of the bullet into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to gravity.
Given:
Initial velocity (V) = 300 m/s
Angle (θ) = 30 degrees
To find the horizontal component (Vx):
Vx = V * cos(θ)
Vx = 300 * cos(30)
Vx ≈ 300 * 0.866
Vx ≈ 259.8 m/s (rounded to one decimal place)
To find the vertical component (Vy):
Vy = V * sin(θ)
Vy = 300 * sin(30)
Vy = 300 * 0.5
Vy = 150 m/s
Next, we need to determine the time it takes for the bullet to strike the ground. We'll use the equation:
t = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
t = (2 * 150) / 9.8
t ≈ 30.6 s (rounded to one decimal place)
Finally, we can calculate the distance traveled horizontally using the equation:
d = Vx * t
d = 259.8 * 30.6
d ≈ 7947.8 m (rounded to one decimal place)
Therefore, the bullet will strike the ground at a distance of approximately 7947.8 meters from the gun.