The initial kinetic energy imparted to a 0.72 kg bullet is 1777 J.The acceleration of gravity is 9.81 m/s2. Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained.
Answer in units of km.
To find the range of the projectile when it is fired at an angle such that the range equals the maximum height attained, we need to use the equations of motion.
First, let's find the maximum height reached by the projectile. At the maximum height, the vertical component of velocity becomes zero. We can use the following equation to find the time taken to reach the maximum height:
vf = vi + at
Where:
vf = final velocity (0 m/s at maximum height)
vi = initial velocity (vertical component of velocity at launch)
a = acceleration due to gravity (-9.81 m/s2)
t = time taken to reach the maximum height (unknown)
Rearranging the equation, we have:
t = (vf - vi) / a
Since vf = 0, the equation becomes:
t = -vi / a
Now, let's find the range of the projectile. The range is the horizontal distance covered by the projectile and depends on the time of flight (total time in the air).
We can use the following equation to find the total time of flight:
t_total = 2t
Where:
t_total = total time of flight (unknown)
t = time taken to reach the maximum height (already found)
Now, let's substitute the value of t from the previous step into the equation for t_total:
t_total = 2 * (-vi / a)
Next, we can use the following equation to find the range:
range = horizontal component of velocity * t_total
The horizontal component of velocity can be determined from the initial kinetic energy of the bullet. The kinetic energy is given by:
KE = (1/2) * m * v^2
Where:
KE = Initial kinetic energy (1777 J in this case)
m = mass of the bullet (0.72 kg)
v = velocity of the projectile (unknown)
Rearranging the equation, we have:
v = sqrt((2 * KE) / m)
Plugging in the given values, we can solve for the velocity.
Once we have the value of v, we can substitute it into the equation for the range:
range = v_horizontal * t_total
Finally, we need to convert the range from meters to kilometers.
So, to summarize the steps:
1. Calculate the time taken to reach the maximum height using t = -vi / a, where vf = 0.
2. Calculate the total time of flight using t_total = 2 * t.
3. Calculate the horizontal component of the velocity using v = sqrt((2 * KE) / m).
4. Calculate the range using range = v_horizontal * t_total.
5. Convert the range from meters to kilometers, if necessary.
To find the range of the projectile when it is fired at an angle such that the range equals the maximum height attained, we need to use the principles of projectile motion.
Let's start by finding the initial velocity of the bullet. We know that kinetic energy is given by the equation:
KE = (1/2) * mass * velocity^2
Given that the mass of the bullet is 0.72 kg and the initial kinetic energy is 1777 J, we can rearrange the equation to solve for the velocity:
velocity^2 = (2 * KE) / mass
velocity^2 = (2 * 1777 J) / 0.72 kg
velocity^2 = 4919.44 J/kg
velocity = √(4919.44) m/s
velocity ≈ 70.09 m/s (rounded to two decimal places)
Now, let's find the angle at which the projectile should be fired to achieve the maximum height. At the maximum height, the vertical component of the initial velocity is zero. We can find this angle using the formula:
tan(theta) = (vertical velocity) / (horizontal velocity)
Since the vertical velocity is zero at the maximum height, we have:
tan(theta) = 0 / (horizontal velocity)
tan(theta) = 0 / velocity
tan(theta) = 0
theta = arctan(0)
theta = 0 degrees
Therefore, the projectile should be fired at an angle of 0 degrees (horizontally) to achieve the maximum height.
Now, let's find the range of the projectile. The range is the horizontal distance traveled by the projectile before hitting the ground. It can be calculated using the formula:
range = (initial velocity)^2 * sin(2 * theta) / acceleration due to gravity
Given that the initial velocity is 70.09 m/s, the angle is 0 degrees, and the acceleration due to gravity is 9.81 m/s^2, we can plug in these values into the formula:
range = (70.09 m/s)^2 * sin(2 * 0 degrees) / 9.81 m/s^2
range = (4909.8081 m^2/s^2) * sin(0 degrees) / 9.81 m/s^2
range = (4909.8081 m^2/s^2) * 0 / 9.81 m/s^2
range = 0 m
Therefore, the range of the projectile when it is fired at an angle such that the range equals the maximum height attained is 0 meters.