A stone is thrown vertically upward with a speed of 11.0 m/s from the edge of a cliff 80.0 m high. How much later does it reach the bottom of the cliff?
v=u+at
0=11.0m/s+(-9.81)t
11.0/9.81=t
t=1.121304791
total time
t=2t
t=2*1.121304791
t=2.242609582
t=2.24s
To solve this problem, we can use the kinematic equations of motion. In this case, we know the initial velocity (u), the height (h), and we need to find the time (t) it takes for the stone to reach the bottom of the cliff.
The equation we can use is:
h = ut + (1/2)gt^2
Where:
- h is the height (80.0 m)
- u is the initial velocity (11.0 m/s)
- g is the acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
- t is the time
We can rearrange this equation to solve for t:
t = (sqrt(2h/g) - u/g)
Now we can substitute in the given values:
t = (sqrt(2 * 80.0 / -9.8) - 11.0 / -9.8)
Calculating this expression gives us:
t ≈ 5.18 seconds
Therefore, the stone takes approximately 5.18 seconds to reach the bottom of the cliff.