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?
h= gt²/2
t=sqrt(2h/g) = sqrt(2•196/9.8)=6.32 s.
The first 96 m
h/2=gt₁²/2
t₁=sqrt(h/g)=sqrt(196/9.8)=4.47 s.
The 2nd 96 m
t₂=t-t₁=6.32-4.47 =1.85 s
To answer these questions, we can use the equations of motion for free fall.
(a) To find the time it takes for the rock to fall the first 98.0 m, we need to find the time it takes for the rock to travel that distance.
The equation that relates distance, initial velocity, time, and acceleration for free fall is:
d = v0 * t + (1/2) * a * t^2
where:
- d is the distance (98.0 m)
- v0 is the initial velocity (0 m/s when the rock is dropped)
- t is the time we want to find
- a is the acceleration due to gravity (-9.8 m/s^2)
Rearranging the equation, we get:
t^2 + (2 * v0 / a) * t - (2 * d / a) = 0
Plugging in the values, we have:
t^2 - 2 * (0 / 9.8) * t - (2 * 98 / 9.8) = 0
Simplifying further, we have:
t^2 - 20t - 20 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -20, and c = -20. Plugging these values into the quadratic formula, we get:
t = (-(-20) ± √((-20)^2 - 4 * 1 * (-20))) / 2 * 1
Simplifying further, we have:
t = (20 ± √(400 + 80)) / 2
t = (20 ± √480) / 2
Now, let's calculate the time it takes for the rock to fall the first 98.0 m:
t = (20 ± √480) / 2
t ≈ 4.9 s (approximation)
(b) Similarly, to find the time it takes for the rock to fall the second 98.0 m, we use the same equation of motion.
Plugging in the values, we have:
t^2 - 2 * (0 / 9.8) * t - (2 * 98 / 9.8) = 0
Simplifying further, we have:
t^2 - 20t - 20 = 0
Using the quadratic formula, we get:
t = (20 ± √(400 + 80)) / 2
t = (20 ± √480) / 2
Let's calculate the time it takes for the rock to fall the second 98.0 m:
t = (20 ± √480) / 2
t ≈ 4.9 s (approximation)
Therefore, both the first and second 98.0 m distances are covered by the rock in approximately 4.9 seconds.