A balloon rising vertically with a velocity of 5m/s releases an object from a height of 100m.(a)How many seconds after it's release will the object reach the ground?(b)At what velocity will the object reach the ground?
height = -4.9t^2 + 5t + 100
a) you want height = 0
-4.9t^2 + 5t + 100 = 0
4.9t^2 - 5t - 100 = 0
solve for t, reject the negative answer
b) d(height)/dt = -9.8t + 5
sub in your answer from a)
asnakc
To solve this problem, we can use the equations of motion. We will consider the downward direction as positive. Let's break it down step-by-step:
(a) We can use the equation of motion:
s = ut + (1/2) * a * t^2,
where:
s = displacement (in this case, -100m, as it is downward)
u = initial velocity of the object (0m/s, as it is initially released)
t = time taken
a = acceleration (which is equal to the acceleration due to gravity, approximately 9.8m/s^2)
Applying the values to the equation, we have:
-100 = 0 * t + (1/2) * (-9.8) * t^2
-100 = (-4.9) * t^2
Divide by -4.9 on both sides:
t^2 = 100 / 4.9
t^2 = 20.41
Taking the square root of both sides, we get:
t = √20.41
t ≈ 4.52 seconds
Therefore, the object will reach the ground approximately 4.52 seconds after its release.
(b) To find the velocity at which the object reaches the ground, we can use the equation of motion:
v = u + a * t,
where:
v = final velocity (which is the velocity at which the object reaches the ground)
u = initial velocity of the object (0m/s, as it is initially released)
t = time taken (which is approximately 4.52 seconds)
a = acceleration (which is equal to the acceleration due to gravity, approximately 9.8m/s^2)
Applying the values to the equation, we have:
v = 0 + 9.8 * 4.52
v ≈ 44.5 m/s
Therefore, the object will reach the ground with a velocity of approximately 44.5 m/s.
To solve this problem, we can use the equations of motion under constant acceleration.
(a) To find the time it takes for the object to reach the ground, we will use the second equation of motion:
\(S = ut + \frac{1}{2}at^2\)
Where:
- \(S\) is the displacement (change in height)
- \(u\) is the initial velocity
- \(a\) is the acceleration (in this case, the acceleration due to gravity, which is approximately \(9.8 m/s^2\))
- \(t\) is the time we want to find
Given:
- Initial velocity, \(u = 0\) (since the object is released)
- Displacement, \(S = -100\) (negative because the object is moving downward)
- Acceleration, \(a = 9.8 m/s^2\)
Plugging in these values into the equation, we get:
\(-100 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2\)
This simplifies to:
\(4.9t^2 = 100\)
Dividing both sides by 4.9:
\(t^2 = \frac{100}{4.9}\)
Taking the square root of both sides:
\(t = \sqrt{\frac{100}{4.9}}\)
Using a calculator, we find:
\(t \approx 4.04\) seconds
Therefore, after approximately 4.04 seconds, the object will reach the ground.
(b) To find the velocity of the object when it reaches the ground, we can use the first equation of motion:
\(v = u + at\)
Where:
- \(v\) is the final velocity
- \(u\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is the time taken (which we just found to be 4.04 seconds)
Given:
- Initial velocity, \(u = 0\) (since the object is released)
- Acceleration, \(a = 9.8 m/s^2\)
- Time, \(t = 4.04\) seconds
Plugging in these values into the equation, we get:
\(v = 0 + 9.8 \cdot 4.04\)
Simplifying this, we find:
\(v \approx 39.59\) m/s
Therefore, the object will reach the ground with a velocity of approximately 39.59 m/s.