A rock is dropped from a 196-m-high cliff. How long does it take to fall (a) the first 98.0 m and (b) the second 98.0 m?
a. d = Vo*t + 0.5g*t^2 = 98 m.
0 + 4.9t^2 = 98
t^2 = 20
Tf = 4.47 = Fall time.
V = Vo + g*t = 0 + 9.8*4.47 = 43.8 m/s.=
Velocity @ 98 m.
b. d = Vo*t + 0.5g*t^2 = 98 m.
43.8*t + 4.9t^2 = 98
4.9t^2 + 43.8t - 98 = 0
Tf = 1.85 s.
To find the time it takes for the rock to fall a certain distance, we can use the equations of motion. In this case, we will use the equation for the time it takes to fall a certain distance h:
h = (1/2)gt^2
where h is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
a) To find the time it takes for the rock to fall the first 98.0 m, we can substitute h = 98.0 m into the equation:
98.0 = (1/2)(9.8)t^2
Simplifying the equation, we get:
t^2 = (2 * 98.0) / 9.8
t^2 = 20.0
t ≈ √20.0
t ≈ 4.47 seconds
Therefore, it takes approximately 4.47 seconds for the rock to fall the first 98.0 m.
b) To find the time it takes for the rock to fall the second 98.0 m, we can subtract the time it took to fall the first 98.0 m from the total time it took to fall 196 m.
Since we already calculated that it took 4.47 seconds to fall the first 98.0 m, we subtract this time from the total time.
Total time = time for the first 98.0 m + time for the second 98.0 m
Total time ≈ 4.47 seconds + 4.47 seconds
Total time ≈ 8.94 seconds
Therefore, it takes approximately 8.94 seconds for the rock to fall the second 98.0 m.
To find the time it takes for the rock to fall for a certain distance, we can use the kinematic equation for uniform acceleration:
\[ d = v_0 t + \frac{1}{2} a t^2 \]
Where:
- \( d \) is the distance traveled
- \( v_0 \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time taken
In this case, the rock is falling freely under gravity, so the initial velocity, \( v_0 \), is 0 m/s. The acceleration, \( a \), is the acceleration due to gravity, \( g \), which is approximately 9.8 m/s\(^2\) (assuming no air resistance).
(a) For the first 98.0 m of fall, we need to find the time taken. Plugging in the given values into the equation, we have:
\[ 98.0 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Simplifying the equation, we get:
\[ 4.9 t^2 = 98.0 \]
Dividing both sides by 4.9, we have:
\[ t^2 = 20.0 \]
Taking the square root of both sides, we find:
\[ t \approx 4.47 \text{ s} \]
Therefore, it takes approximately 4.47 seconds for the rock to fall the first 98.0 m.
(b) For the second 98.0 m of fall, we can use the same equation since the acceleration remains the same. We need to find the time taken for this distance:
\[ 98.0 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Following the same steps as before, we find:
\[ t \approx 4.47 \text{ s} \]
Again, it takes approximately 4.47 seconds for the rock to fall the second 98.0 m.
Therefore, the time taken for both (a) the first 98.0 m and (b) the second 98.0 m is approximately 4.47 seconds.