A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 85.0 m high (Fig. 2-34).
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?
It is the same equations as always:
d=Vi*t-1/2 g t^2
Vf=Vi+gt
Now for total distance, you will have to calculate the height it reaches at the top (when vf is zero). Double that, add to height of cliff.
5.05s
To answer these questions, we need to use the equations of motion for an object in free fall. We'll use the following equations:
1. Vertical displacement (h) = initial velocity (u) x time (t) + 0.5 x acceleration due to gravity (g) x time squared (t^2)
2. Final velocity (v) = initial velocity (u) + acceleration due to gravity (g) x time (t)
3. Final velocity squared (v^2) = initial velocity squared (u^2) + 2 x acceleration due to gravity (g) x vertical displacement (h)
Given:
Initial velocity (u) = 12.0 m/s (upward)
Height of the cliff (h) = 85.0 m
Acceleration due to gravity (g) = 9.8 m/s² (downward)
(a) To find the time it takes for the stone to reach the bottom of the cliff, we need to find the time it takes for the stone to reach its maximum height first. At the maximum height, the final velocity will be zero.
Using equation 2:
0 = 12.0 m/s - 9.8 m/s² x t_max
9.8 m/s² x t_max = 12.0 m/s
t_max = 12.0 m/s / 9.8 m/s²
t_max = 1.22 s
Now, we can use this value of t_max to find the total time it takes for the stone to reach the bottom of the cliff:
t_total = 2 x t_max
t_total = 2 x 1.22 s
t_total = 2.44 s
Therefore, the stone takes 2.44 seconds to reach the bottom of the cliff.
(b) To find the speed of the stone just before hitting the ground, we can use equation 2 again:
v = 12.0 m/s - 9.8 m/s² x t_total
v = 12.0 m/s - 9.8 m/s² x 2.44 s
v = 12.0 m/s - 23.87 m/s
v ≈ -11.87 m/s
The negative sign indicates that the velocity is in the opposite direction of the initial velocity, which means the stone is moving downward.
Therefore, the speed of the stone just before hitting the ground is approximately 11.87 m/s.
(c) To find the total distance traveled by the stone, we need to calculate the distance traveled during the upward journey and the distance traveled during the downward journey.
During the upward journey, the stone travels the vertical displacement (height of the cliff) in time t_max. Using equation 1:
Distance_upward = 12.0 m/s x t_max + 0.5 x (-9.8 m/s²) x t_max²
Distance_upward = 12.0 m/s x 1.22 s + 0.5 x (-9.8 m/s²) x (1.22 s)²
Distance_upward ≈ 14.63 m
During the downward journey, the stone travels the vertical displacement (height of the cliff) in time t_total - t_max. Using equation 1 again:
Distance_downward = 0 m/s x (t_total - t_max) + 0.5 x (-9.8 m/s²) x (t_total - t_max)²
Distance_downward = 0 m/s x (2.44 s - 1.22 s) + 0.5 x (-9.8 m/s²) x (2.44 s - 1.22 s)²
Distance_downward ≈ -57.17 m
The negative sign represents the downward motion of the stone.
To find the total distance traveled, we take the sum of the upward distance and the absolute value of the downward distance:
Total_distance = |Distance_upward| + |Distance_downward|
Total_distance = |14.63 m| + |-57.17 m|
Total_distance = 14.63 m + 57.17 m
Total_distance ≈ 71.80 m
Therefore, the total distance traveled by the stone is approximately 71.80 meters.