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A rifle that shoots bullets at [tex]$460 \, \text{m/s}$[/tex] is to be aimed at a target [tex]$45.7 \, \text{m}$[/tex] away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

Answer :

The rifle barrel must be pointed 0.1313 m (or 13.13 cm) above the target to hit dead center. To determine how high above the target the rifle barrel must be pointed so that the bullet hits dead center, we need to use the formula for projectile motion:

y = x*tan(θ) - (g*x^2)/(2*v^2*cos^2(θ))

where:
y = vertical displacement (how high above the target the rifle barrel must be pointed)
x = horizontal displacement (distance to the target)
θ = angle of elevation of the rifle barrel
v = initial velocity of the bullet
g = acceleration due to gravity

Given:
x = 45.7 m
v = 460 m/s
g = 9.8 m/s^2

We need to find y and θ.

To find θ, we can rearrange the formula and use the quadratic formula:

0 = x*tan(θ) - (g*x^2)/(2*v^2*cos^2(θ))

0 = tan(θ) - (g*x)/(2*v^2*cos^2(θ))

0 = tan(θ)*cos^2(θ) - (g*x)/(2*v^2)

0 = sin(θ)*cos(θ) - (g*x)/(2*v^2)

0 = 2*sin(θ)*cos(θ) - (g*x)/v^2

0 = sin(2*θ) - (g*x)/v^2

sin(2*θ) = (g*x)/v^2

2*θ = arcsin((g*x)/v^2)

θ = (1/2)*arcsin((g*x)/v^2)

θ = (1/2)*arcsin((9.8*45.7)/(460^2))

θ = 0.00287 radians

Now that we have θ, we can plug it back into the formula to find y:

y = x*tan(θ) - (g*x^2)/(2*v^2*cos^2(θ))

y = 45.7*tan(0.00287) - (9.8*45.7^2)/(2*460^2*cos^2(0.00287))

y = 0.1313 m

To learn more about projectile motion here:

https://brainly.com/question/24216590#

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Rewritten by : Jeany

Final answer:

To hit the target dead center, the rifle barrel must be pointed approximately 33.3 meters above the target.

Explanation:

To determine how high above the target the rifle barrel must be aimed, we need to consider the time it takes for the bullet to reach the target and the effect of gravity on the bullet's trajectory. Since the bullet is fired horizontally, it will take the same time to reach the target as it would if it was dropped vertically from the height of the rifle barrel. Using the equation h = (1/2)gt^2, where h is the vertical displacement, g is the acceleration due to gravity (9.8 m/s^2), and t is the time, we can solve for t. Plugging in the values, we have:



h = (1/2)(9.8)(t^2)

45.7 = (1/2)(9.8)(t^2)

t^2 = (2 × 45.7) / 9.8

t = √(2 × 45.7) / √9.8

t = 3.4 seconds (rounded to one decimal place)



Since the bullet takes 3.4 seconds to reach the target, the barrel must be aimed high enough so that the bullet's vertical displacement is zero at this time. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can solve for the initial vertical velocity of the bullet. Plugging in the values, we have:



0 = u + (9.8)(3.4)

u = - (9.8)(3.4)

u = -33.3 m/s (rounded to one decimal place)



Therefore, the rifle barrel must be pointed approximately 33.3 meters above the target in order for the bullet to hit dead center.