A parachutist with a camera, both descending at a speed of 13 m/s, releases the camera
at an altitude of 57.8 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.
Answer in units of m/s
How long does it take the camera to reach the
ground?
Answer in units of s
Why did the parachutist release the camera? Because it wanted to capture the "ground-breaking" moment!
Now, let's solve this "falling camera" mystery.
To find the magnitude of the camera's velocity just before it hits the ground, we can use the equation:
(vf)² = (vi)² + 2ad
Where:
vf = final velocity
vi = initial velocity (which is 13 m/s, same as the parachutist's)
a = acceleration due to gravity (9.8 m/s²)
d = distance traveled (57.8 m)
Substituting the values into the equation, we have:
(vf)² = (13)² + 2(-9.8)(57.8)
Calculating that out, we get:
(vf)² = 169 + (-1130.48)
(vf)² = -961.48
Now, since we're only interested in the magnitude (or absolute value) of the velocity, we take the square root of the result:
vf ≈ √(|-961.48|)
vf ≈ √961.48
vf ≈ 31.00 m/s
So, the magnitude of the velocity of the camera just before it hits the ground is approximately 31.00 m/s.
Now, for the time it takes the camera to reach the ground, we can use the equation:
d = vit + 0.5at²
Where, again:
d = distance traveled (57.8 m)
vi = initial velocity (13 m/s)
a = acceleration due to gravity (9.8 m/s²)
t = time
Rearranging the equation and substituting the values, we have:
57.8 = 13t + 0.5(9.8)t²
Simplifying and rearranging it further, we get a quadratic equation:
0.5(9.8)t² + 13t - 57.8 = 0
Solving this equation, we find two possible values for t: t ≈ 2.71 s and t ≈ -4.21 s.
Since time can't be negative in this case, we discard the negative value:
t ≈ 2.71 s
So, it takes approximately 2.71 seconds for the camera to reach the ground.
Just remember, if you ever drop your camera, make sure it's on purpose, and try to capture some "falling memories" on the way down!
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 = distance
In this case, the initial velocity (u) is 13 m/s, the acceleration (a) is 9.8 m/s^2, and the distance (s) is 57.8 m (which is the altitude).
Substituting the values into the equation, we get:
\(v^2 = (13)^2 + 2(-9.8)(-57.8)\)
\(v^2 = 169 + 1130.64\)
\(v^2 = 1299.64\)
Taking the square root of both sides, we find:
\(v = \sqrt{1299.64}\)
\(v \approx 36.03\) m/s
So, the magnitude of the velocity of the camera just before it hits the ground is approximately 36.03 m/s.
To find how long it takes the camera to reach the ground, we can use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Where:
s = distance (which is the altitude = 57.8 m)
u = initial velocity (13 m/s)
a = acceleration (9.8 m/s^2)
t = time
Substituting the values into the equation, we get:
\(57.8 = (13)t + \frac{1}{2}(9.8)(t^2)\)
\(57.8 = 13t + 4.9t^2\)
Rearranging the equation:
\(4.9t^2 + 13t - 57.8 = 0\)
To solve this quadratic equation, we can either factorize it or use the quadratic formula. In this case, the quadratic formula will be used:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substituting the values into the formula, we get:
\(t = \frac{-13 \pm \sqrt{13^2 - 4(4.9)(-57.8)}}{2(4.9)}\)
Calculating the value under the square root:
\(\sqrt{13^2 - 4(4.9)(-57.8)} \approx 38.18\)
Substituting this value into the formula:
\(t = \frac{-13 \pm 38.18}{2(4.9)}\)
Calculating the two possible values of time:
\(t_1 = \frac{-13 + 38.18}{2(4.9)} \approx 2.30\) s
\(t_2 = \frac{-13 - 38.18}{2(4.9)} \approx -2.91\) s
Since time cannot be negative in this context, we discard the negative value. Therefore, the time it takes the camera to reach the ground is approximately 2.30 s.
To find the magnitude of the velocity of the camera just before it hits the ground, we can use the equations of motion. Since air friction is negligible, the only force acting on the camera is gravity. The initial velocity when the camera is released is the same as the parachutist's speed of 13 m/s.
Using the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration (in this case, due to gravity)
s = displacement (in this case, the distance from the release point to the ground)
Substituting the values into the equation:
v^2 = 13^2 + 2(-9.8)(57.8)
v^2 = 169 + 2(-9.8)(57.8)
v^2 = 169 - 1132.24
v^2 = - 963.24
Since velocity cannot be negative, we disregard the negative sign. So, the magnitude of the velocity of the camera just before it hits the ground is:
v = √963.24
v ≈ 31 m/s
Now, to find out how long it takes for the camera to reach the ground, we can use the equation:
v = u + at
where
v = final velocity (which is 31 m/s)
u = initial velocity (which is 13 m/s)
a = acceleration (which is -9.8 m/s^2, negative because it is in the opposite direction of the initial velocity)
t = time
Rearranging the equation, we have:
t = (v - u) / a
t = (31 - 13) / -9.8
t = 18 / -9.8
t ≈ -1.84 s
The negative sign indicates that it takes approximately 1.84 seconds for the camera to hit the ground after it is released. However, since time cannot be negative, we ignore the negative sign.
Therefore, the camera takes approximately 1.84 seconds to reach the ground.