a particle falling under gravity falls 39m in certain second find time required to cover the next 30 m take g=9.8
V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*39 = 764.4
V = 27.65 m/s.
d = Vo*t + 0.5g*t^2 = 30 m.
27.65t + 4.9t^2 = 30
4.9t^2 + 27.65t - 30 = 0
Use Quadratic Formula and get:
Tf = 0.931 s. = Fall time.
To find the time required to cover the next 30 m, we can use the equations of motion under constant acceleration.
Let's consider the first part of the particle's motion, where it falls 39 m. We can use the formula for distance traveled under uniform acceleration:
s = ut + (1/2)at^2
where:
s = distance traveled
u = initial velocity (which is zero because the particle is falling under gravity)
a = acceleration due to gravity (g = 9.8 m/s^2)
t = time taken
Since the initial velocity is zero, the equation simplifies to:
s = (1/2)gt^2
Plugging in the known values of s and g:
39 = (1/2)(9.8)(t^2)
Now, let's solve for t:
39 = 4.9t^2
t^2 = 39/4.9
t^2 = 8
t = √8
t ≈ 2.83 seconds
So, the time taken to cover the first 39 m is approximately 2.83 seconds.
Now, to find the time required to cover the next 30 m, we can use the same formula:
s = ut + (1/2)at^2
Here, the initial velocity (u) is the final velocity of the previously calculated motion, which is the time derivative of the distance (s) with respect to time (t). Let's denote it as v:
v = gt
The second part of the motion covers 30 m, so we have:
30 = vt + (1/2)at^2
Plugging in the known values of v (which is gt) and a:
30 = (gt)t + (1/2)(9.8)t^2
Simplifying the equation:
30 = 9.8t^2 + 4.9t^2
30 = 14.7t^2
Solving for t:
t^2 = 30/14.7
t^2 ≈ 2.04
t ≈ √2.04
t ≈ 1.43 seconds
Therefore, the time required to cover the next 30 m is approximately 1.43 seconds.
To find the time required to cover the next 30 meters after falling 39 meters, we can use the equation for the distance traveled by an object under free-fall acceleration:
s = ut + (1/2)gt^2
Where:
- s is the distance traveled
- u is the initial velocity (which is 0 in this case since the particle is falling)
- t is the time
- g is the acceleration due to gravity
Given that the particle falls 39 meters, we can use this information to find the time it took to fall this distance. Plug in the known values into the equation:
39 = (0)t + (1/2)(9.8)(t^2)
Now, we can solve this quadratic equation for t. Simplify the equation:
39 = (4.9)(t^2)
Divide both sides by 4.9 to isolate t^2:
(t^2) = 39 / 4.9
(t^2) ≈ 7.95918
Finally, take the square root of both sides to find t:
t ≈ √(7.95918)
t ≈ 2.82 seconds (rounded to two decimal places)
Therefore, it took approximately 2.82 seconds for the particle to fall 39 meters.
To find the time required to cover the next 30 meters, we can use the same equation, but this time the initial velocity will not be zero, as the particle has already fallen 39 meters. The equation becomes:
s = ut + (1/2)gt^2
We need to solve for t when the distance traveled (s) is 30 meters. Plug in the values into the equation:
30 = (0)t + (1/2)(9.8)(t^2)
Simplify the equation:
15t^2 = 30
Divide both sides by 15:
t^2 = 2
Take the square root of both sides:
t ≈ √2
t ≈ 1.41 seconds (rounded to two decimal places)
Therefore, it would take approximately 1.41 seconds to cover the next 30 meters.