A camera falls from a blimp that is 399 m above the ground and rising at a speed of 9.75 m/s. Find the maximum height reached by the camera with respect to the ground. The acceleration of gravity is 9.8 m/s^2.Answer in units of m.
b)Find the speed at which it hits the ground.Answer in units of cm/s.
Need Help ASAP please in the next 15minutes!!!
Use kinematic equations,velocity at max height is zero, so 0=9.75^2-19.6(X-399)=403.9m. Then for speed at ground just do V^2=-19.6(0-403)=88.9 m/s
ahhh thanks but just a tad late lol, due already
To find the maximum height reached by the camera, we can first calculate the time it takes for the camera to reach its highest point.
We can use the equation of motion:
h = h0 + v0t + (1/2)gt^2
Where:
h = final height
h0 = initial height
v0 = initial velocity
g = acceleration due to gravity
t = time
Since the camera is falling downwards, the initial velocity is negative. The initial height is 399 m, the initial velocity is -9.75 m/s (since it's falling downward), and the acceleration due to gravity is -9.8 m/s^2.
At the highest point, the final height will be the maximum height reached. The final velocity will be 0 m/s since the camera momentarily comes to rest at its highest point.
Using these values, we can rearrange the equation to solve for t:
0 = 399 - 9.75t + (1/2)(-9.8)t^2
Multiply out the equation:
-4.9t^2 - 9.75t + 399 = 0
This is a quadratic equation. We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac))/(2a)
Where a = -4.9, b = -9.75, and c = 399.
Plugging in these values, we get:
t = (-(-9.75) ± √((-9.75)^2 - 4(-4.9)(399)))/(2(-4.9))
Simplifying further, we have:
t = (9.75 ± √(95.0625 + 7836))/(9.8)
t = (9.75 ± √7931.0625)/(9.8)
We only consider the positive root since time cannot be negative in this context. Therefore, the time taken to reach the maximum height is:
t = (9.75 + √7931.0625)/(9.8) ≈ 9.020 seconds
To find the maximum height reached by the camera, we can substitute this value of t back into the equation of motion:
h = h0 + v0t + (1/2)gt^2
h = 399 - 9.75(9.020) + (1/2)(-9.8)(9.020)^2
h ≈ 399 - 87.978 + 0.5(-9.8)(81.3604)
h ≈ 399 - 87.978 - 400.2
h ≈ -88.178 meters
Since the height cannot be negative, we can take the absolute value of this result to find the maximum height reached by the camera:
Maximum height reached by the camera = |-88.178| ≈ 88.178 meters.
Now, to find the speed at which it hits the ground, we can use the equation:
v = v0 + gt
Where:
v = final velocity
v0 = initial velocity
g = acceleration due to gravity
t = time
At the time the camera hits the ground, the final velocity is the speed at which it hits the ground. The initial velocity is -9.75 m/s and the acceleration due to gravity is -9.8 m/s^2.
Using these values, we can substitute them into the equation to find the final velocity:
v = -9.75 + (-9.8)(9.020)
v ≈ -9.75 - 88.1776
v ≈ -97.9276 m/s
To convert this to cm/s, we can multiply by 100:
Final Velocity = -97.9276 m/s * 100 = -9792.76 cm/s
Therefore, the speed at which it hits the ground is approximately -9792.76 cm/s.
Note: The negative sign indicates that the velocity is downward.
To find the maximum height reached by the camera, we can use the kinematic equation for vertical motion:
h = h₀ + v₀t + (1/2)at²
Where:
h is the final height
h₀ is the initial height
v₀ is the initial velocity
t is the time
a is the acceleration
In this case, the initial height is 399 m, the initial velocity is 9.75 m/s (upwards), the acceleration is -9.8 m/s² (downwards due to gravity), and we want to find the time it takes to reach the maximum height. At the maximum height, the velocity is 0 m/s.
Using the equation v = v₀ + at, we can find the time it takes to reach the maximum height:
0 = 9.75 - 9.8t
9.8t = 9.75
t = 9.75 / 9.8
t ≈ 0.9949 s
Now, we can substitute this time into the equation for height to find the maximum height:
h = 399 + 9.75 * 0.9949 + (1/2) * (-9.8) * (0.9949)²
h ≈ 399 + 9.7039 + (-4.902) ≈ 403.801 m
Therefore, the maximum height reached by the camera with respect to the ground is approximately 403.801 m.
To find the speed at which it hits the ground, we can use the equation:
v = v₀ + at
Where:
v is the final velocity
v₀ is the initial velocity
t is the time
a is the acceleration
In this case, the initial velocity is 9.75 m/s (upwards), the acceleration is -9.8 m/s² (downwards due to gravity), and we want to find the final velocity when the camera hits the ground. The final velocity is what we need to find.
Using the equation v = v₀ + at, we can find the final velocity:
v = 9.75 + (-9.8) * t
Substituting the value of t we found earlier:
v ≈ 9.75 + (-9.8) * 0.9949
v ≈ 9.75 - 9.704102 ≈ 0.045 cm/s
Therefore, the speed at which the camera hits the ground is approximately 0.045 cm/s.