A bullet is fired horizontally with a velocity of 250 m/s from the top of a building 100m high. Find the Horizontal and vertical velocities of the bullet when it reaches the ground, the resultant velocity, the horizontal displacement, the net displacement, the direction of the net displacement.
horizontal velocity remains constant throughout the flight. There is no horizontal force or acceleration. So 250 m/s at top and at ground
Now the vertical problem, which is simply falling from 100 meters
-100 = 0 + 0 t - 4.9 t^2
t = 4.52 seconds to fall
v
down speed at ground = -9.8 t = -44.3 m/s
so speed at ground is 250 horizontal and - 44.3 vertical. You do the trig
It moves 100 meters down and 250*4.52 meters horizontal.
again you do the trig
To solve this problem, we need to consider the motion of the bullet in the horizontal and vertical directions separately.
1. Finding the horizontal velocity (Vx):
Since the bullet is fired horizontally, there will be no change in its horizontal velocity throughout its motion. Therefore, the horizontal velocity of the bullet remains constant at 250 m/s. So, the horizontal velocity (Vx) is 250 m/s.
2. Finding the vertical velocity (Vy) when it reaches the ground:
Since the bullet is only affected by gravity in the vertical direction, we can use the equation of motion:
Vy^2 = Uy^2 + 2ayΔy
where Vy is the final vertical velocity, Uy is the initial vertical velocity (which is 0 m/s since it is fired horizontally), ay is the acceleration due to gravity (which is -9.8 m/s^2), and Δy is the vertical displacement (which is -100 m since the bullet falls downwards).
Plugging in the values:
Vy^2 = 0^2 + 2(-9.8)(-100)
Vy^2 = 9800
Vy = sqrt(9800)
Vy ≈ 99 m/s (rounded to two decimal places)
3. Finding the resultant velocity (Vr):
The resultant velocity is the vector sum of the horizontal and vertical velocities. To find it, we can use the Pythagorean theorem:
Vr^2 = Vx^2 + Vy^2
Vr^2 = 250^2 + 99^2
Vr^2 ≈ 67751
Vr ≈ sqrt(67751)
Vr ≈ 260 m/s (rounded to two decimal places)
4. Finding the horizontal displacement (Hd):
To calculate the horizontal displacement covered by the bullet, we can use the formula:
Hd = Vx × t
where Hd is the horizontal displacement and t is the total time of flight.
To find the time of flight, we need to use the equation:
Δy = Uy × t + (1/2)ay × t^2
where Δy is the vertical displacement (which is -100 m), Uy is the initial vertical velocity (which is 0 m/s), and ay is the acceleration due to gravity (which is -9.8 m/s^2).
Plugging in the values:
-100 = 0 × t + (1/2)(-9.8) × t^2
-100 = -4.9t^2
t^2 = 100/4.9
t ≈ sqrt(20.41)
t ≈ 4.52 s (rounded to two decimal places)
Now, we can calculate the horizontal displacement:
Hd = 250 × 4.52
Hd ≈ 1130 m (rounded to two decimal places)
5. Finding the net displacement:
The net displacement is the vector sum of the horizontal and vertical displacements. Since the bullet starts and ends at the same height, the net displacement is equal to the horizontal displacement. So, the net displacement is approximately 1130 m.
6. Finding the direction of the net displacement:
Since the horizontal displacement is positive (to the right), the direction of the net displacement is also to the right.
To summarize:
- Horizontal velocity (Vx): 250 m/s
- Vertical velocity (Vy) when it reaches the ground: 99 m/s
- Resultant velocity (Vr): approximately 260 m/s
- Horizontal displacement (Hd): approximately 1130 m
- Net displacement: approximately 1130 m
- Direction of the net displacement: to the right