A gun fires a bullet of a velocity of 200m/s at an angle of 40° to the horizontal. Find the velocity of the position of the bullet after 20second. Find also the range and time required of the bullet to reach the ground?
see wikipedia's excellent article on "trajectory"
It gives handy formulas for all of this.
Come back if you get stuck, and show what you got.
To find the velocity of the bullet after 20 seconds, we need to consider the horizontal and vertical components separately.
The horizontal component of the velocity remains constant throughout the motion. As the bullet fired at an angle of 40° to the horizontal, we can calculate the horizontal velocity using trigonometry:
Horizontal velocity = initial velocity * cos(angle)
= 200 m/s * cos(40°)
≈ 200 m/s * 0.766
≈ 153.2 m/s
So, the horizontal component of the velocity remains constant at approximately 153.2 m/s.
Now, let's consider the vertical component of the velocity. The bullet experiences constant acceleration due to gravity, which is approximately -9.8 m/s² (assuming no air resistance).
Using the equation of motion for vertical motion:
Vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
Since the bullet is fired horizontally, the initial vertical velocity is zero.
Vertical displacement = 0 * 20 + (1/2) * -9.8 * (20^2)
= 0 - 9.8 * 200
= -1960 m
Since the initial vertical velocity is zero and the bullet returns to the same height it was fired from, the vertical displacement is -1960 m.
Now, we can calculate the vertical component of the velocity using the equation of motion for final velocity:
Vertical velocity (v) = initial vertical velocity + acceleration * time
= 0 + (-9.8) * 20
= -196 m/s
The negative sign indicates that the bullet is moving downward.
Finally, to find the total velocity of the bullet after 20 seconds, we can combine the horizontal and vertical components using the Pythagorean theorem:
Total velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
= sqrt((153.2)^2 + (-196)^2)
≈ sqrt(23496.24 + 38416)
≈ sqrt(61912.24)
≈ 248.8 m/s
So, after 20 seconds, the velocity of the bullet is approximately 248.8 m/s.
Now, let's find the range and the time required for the bullet to reach the ground.
The range is the horizontal distance traveled by the bullet. To find the range, we can use the equation:
Range = horizontal velocity * time
= 153.2 m/s * 20 s
= 3064 m
So, the range of the bullet is approximately 3064 meters.
To find the time required for the bullet to reach the ground, we can use the equation of motion for vertical motion:
Vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
Since the bullet reaches the ground, the vertical displacement is -1960 m.
-1960 = 0 * time + (1/2) * -9.8 * (time^2)
-1960 = -4.9 * (time^2)
Rearranging the equation:
time^2 = -1960 / -4.9
time^2 = 400
time = sqrt(400)
time = 20 seconds
Therefore, the time required for the bullet to reach the ground is 20 seconds.
To find the velocity of the position of the bullet after 20 seconds, we need to first decompose the initial velocity into its horizontal and vertical components.
The horizontal component of the velocity remains constant throughout the trajectory, so we can find it using the initial velocity (200 m/s) and the angle of projection (40°).
Horizontal component of velocity = Initial velocity * cos(angle)
Horizontal component of velocity = 200 * cos(40°)
Horizontal component of velocity ≈ 200 * 0.766
Horizontal component of velocity ≈ 153.2 m/s
Now, to find the vertical component of the velocity after 20 seconds, we need to take into account the gravitational acceleration acting in the downward direction.
Vertical component of velocity = Initial velocity * sin(angle) - (acceleration due to gravity * time)
Vertical component of velocity = 200 * sin(40°) - (9.8 * 20)
Vertical component of velocity ≈ 200 * 0.6428 - (9.8 * 20)
Vertical component of velocity ≈ 128.56 - 196
Vertical component of velocity ≈ -67.44 m/s _(Negative sign indicates that the bullet is moving downward)_
Now, to find the velocity of the position of the bullet after 20 seconds, we can use the Pythagorean theorem:
Velocity = √(Horizontal component of velocity^2 + Vertical component of velocity^2)
Velocity = √(153.2^2 + (-67.44)^2)
Velocity ≈ √(23500 + 4543.73)
Velocity ≈ √28043.73
Velocity ≈ 167.58 m/s
So, the velocity of the position of the bullet after 20 seconds is approximately 167.58 m/s.
Next, to find the range, we need to determine the horizontal displacement of the bullet when it reaches the ground.
The time required for the bullet to reach the ground can be found by determining the time it takes for the bullet to reach its maximum height and then doubling that time. At the maximum height, the vertical component of velocity becomes zero.
Time to reach maximum height = (Vertical component of velocity) / (acceleration due to gravity)
Time to reach maximum height = (-67.44) / (-9.8)
Time to reach maximum height ≈ 6.87 seconds
Total time of flight = 2 * Time to reach maximum height
Total time of flight = 2 * 6.87
Total time of flight ≈ 13.74 seconds
Now, we can find the range by multiplying the horizontal component of velocity by the total time of flight.
Range = Horizontal component of velocity * Total time of flight
Range ≈ 153.2 * 13.74
Range ≈ 2107.25 meters
So, the range of the bullet is approximately 2107.25 meters.
Finally, the time required for the bullet to reach the ground is the total time of flight, which is 13.74 seconds.