An artillery shell is launched on a flat, horizontal field at an angle of a = 43.5° with respect to the horizontal and with an initial speed of v0 = 263 m/s. What is the horizontal distance covered by the shell after 6.77 s of flight?
What is the height of the shell at this moment?
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 31.0 m above sea level, directed at an angle theta = 48.5° above the horizontal, and with a speed v = 35.6 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.
Suppose that a senior driving a Pontiac zooms out of a darkened tunnel at 31.0 m/s. She is momentarily blinded by the sunshine. When she recovers, she sees that she is fast overtaking a bus ahead in her lane moving at the slower speed of 15.7 m/s. She hits the brakes as fast as she can (her reaction time is 0.40 s). If she can decelerate at 2.5 m/s2, what is the minimum distance between the driver and the bus when she first sees it so that they do not collide?
To find the horizontal distance covered by the shell, we can use the horizontal component of the velocity.
Step 1: Find the horizontal component of the velocity (Vx):
Vx = v0 * cos(a)
Vx = 263 m/s * cos(43.5°)
Vx ≈ 187.6 m/s
Step 2: Find the horizontal distance covered (Dx):
Dx = Vx * t
Dx = 187.6 m/s * 6.77 s
Dx ≈ 1270.1 m
Therefore, the horizontal distance covered by the shell after 6.77 seconds of flight is approximately 1270.1 meters.
To find the height of the shell at this moment, we can use the vertical component of the velocity.
Step 1: Find the vertical component of the velocity (Vy):
Vy = v0 * sin(a)
Vy = 263 m/s * sin(43.5°)
Vy ≈ 178.9 m/s
Step 2: Find the height (H):
H = Vy * t - (0.5 * g * t^2)
where g is the acceleration due to gravity (~9.8 m/s^2)
H = 178.9 m/s * 6.77 s - (0.5 * 9.8 m/s^2 * (6.77 s)^2)
H ≈ 644.1 m
Therefore, the height of the shell after 6.77 seconds of flight is approximately 644.1 meters.
To find the horizontal distance covered by the shell, we can use the horizontal component of the initial velocity and the time of flight.
The horizontal component of the initial velocity (v0x) can be found using the formula:
v0x = v0 * cos(a)
Where:
v0 = 263 m/s (initial speed of the shell)
a = 43.5° (angle of launch with respect to the horizontal)
Plugging in these values:
v0x = 263 m/s * cos(43.5°)
Now that we have the horizontal component of the initial velocity, we can use it in the horizontal distance formula:
x = v0x * t
Where:
t = 6.77 s (time of flight)
Plugging in the values:
x = (263 m/s * cos(43.5°)) * 6.77 s
Now, let's calculate the horizontal distance covered by the shell.
To find the height of the shell at this moment, we can use the vertical component of the initial velocity, the time of flight, and the acceleration due to gravity.
The vertical component of the initial velocity (v0y) can be found using the formula:
v0y = v0 * sin(a)
Plugging in the values:
v0y = 263 m/s * sin(43.5°)
To find the height (y) at this moment, we can use the formula:
y = v0y * t - (1/2) * g * t^2
Where:
g = 9.8 m/s² (acceleration due to gravity)
Plugging in the values:
y = (263 m/s * sin(43.5°)) * 6.77 s - (1/2) * 9.8 m/s² * (6.77 s)^2
Now, let's calculate the height of the shell at this moment.