A parachutist with a camera, both descending at a speed of 14 m/s, releases the camera
at an altitude of 31.7 m.
What is the magitude of the velocity of the
camera just before it hits the ground? The
acceleration of gravity is 9.8 m/s
2
and air
friction is negligible.
Well, it seems like the parachutist decided to give the camera some "sky time," huh? Anyway, let's crunch some numbers here.
Since the parachutist and the camera are both descending at a speed of 14 m/s, we don't need to consider their initial velocities separately. The velocity of the camera just before it hits the ground will be the same as the parachutist's velocity.
Now, in order to find the final velocity of the camera, we need to use the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity
vi is the initial velocity
a is the acceleration (in this case, acceleration due to gravity)
d is the displacement (in this case, the distance from the release point to the ground)
Since air friction is negligible, the acceleration doesn't change. So, we have:
vf^2 = (14 m/s)^2 + 2 * 9.8 m/s^2 * 31.7 m
Now, let's calculate that:
vf^2 = 196 m^2/s^2 + 2 * 9.8 m/s^2 * 31.7 m
vf^2 = 196 m^2/s^2 + 622.36 m^2/s^2
vf^2 = 818.36 m^2/s^2
Taking the square root of both sides:
vf = √818.36 m^2/s^2
Calculating that, we get:
vf ≈ 28.63 m/s
So, the magnitude of the velocity of the camera just before it hits the ground is approximately 28.63 m/s.
To find the magnitude of the velocity of the camera just before it hits the ground, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, the initial velocity (u) is 14 m/s, the acceleration (a) is -9.8 m/s^2 (negative because it's in the opposite direction of the motion), and the displacement (s) is 31.7 m.
Let's substitute the values into the equation to find the final velocity (v):
v^2 = 14^2 + 2(-9.8)(31.7)
v^2 = 196 + (2)(-9.8)(31.7)
v^2 = 196 - 615.32
v^2 = -419.32
Since velocity cannot be negative, we discard the negative value.
Taking the square root of v^2, we get:
v = √(-419.32)
v ≈ 20.47 m/s
Therefore, the magnitude of the velocity of the camera just before it hits the ground is approximately 20.47 m/s.
To find the magnitude of the velocity of the camera just before it hits the ground, we can use the equations of motion.
First, we can determine the time it takes for the camera to hit the ground. We can use the equation:
y = y0 + v0yt + (1/2)gt^2
Where:
y = final position (0, since it hits the ground)
y0 = initial position (31.7 m)
v0y = initial vertical velocity (14 m/s, since both the parachutist and camera are descending at the same speed)
g = acceleration due to gravity (-9.8 m/s^2)
Rearranging the equation to solve for time (t):
y - y0 = v0yt + (1/2)gt^2
0 - 31.7 = (14)t + (1/2)(-9.8)(t^2)
-31.7 = 14t - 4.9t^2
This is a quadratic equation, so we can solve it by rearranging it into the standard form (ax^2 + bx + c = 0) and using the quadratic formula:
4.9t^2 - 14t - 31.7 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
t = (-(-14) ± √((-14)^2 - 4(4.9)(-31.7))) / (2 * 4.9)
Simplifying:
t ≈ 2.22 s
Now that we know the time it takes for the camera to hit the ground, we can find the final velocity using the equation:
v = v0 + gt
Where:
v = final velocity
v0 = initial velocity (14 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (2.22 s)
Plugging in the values:
v = 14 + (-9.8)(2.22)
v ≈ -30.1 m/s
Since the camera is descending, the magnitude of the velocity is the absolute value of the final velocity:
|v| ≈ 30.1 m/s
Therefore, the magnitude of the velocity of the camera just before it hits the ground is approximately 30.1 m/s.