This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop = (initial height) + (initial velocity) x (T) + (1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of the same path that it would have had if it were launched from the ground 5.235 seconds before it was actually dropped from balloon while ascending.)
Concerning the maximum height ... I don't know right now any other easy way to do that part without differentiating the big equation. So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second. So the bag reaches the top of its arc, runs out of gas, and starts falling, after (3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s. At the end of that time, it's moving with zero vertical speed). Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m, then turns and begins to plummet earthward, where it plummets to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion, to express my profound gratitude for the bounty of 5 points that I shall reap from this work. The moldy crust and tepid cloudy water have been delicious, and will not soon be forgotten.
Static equilibrium means that all forces are equal, so make this easiest you want to break F1 into it's horizontal and vertical components. As there are no other forces acting in the horizontal, we know the horizontal component of F1 is 40N. This allows the vertical component to be found using pythagorus theorem. After finding the vertical and horizontal components, you just have to add the vertical components to find the difference between the up and down.
Time period of a pendulum clock is dependent on two factors namely:length and acceleration due to gravity.
When a clock loses time, the time period of the pendulum clock increases.
This however can be corrected by decreasing the length of the pendulum.The time period of the pendulum clock is not dependent on the mass of the bob. The time period of the pendulum clock can be corrected only by changing the length of the pendulum string.
