A stone is thrown from a cliff top with an upward velocity of 20 m/s. It takes 7 seconds for the stone to hit the ground at the bottom of the cliff. How high is the cliff?
I know you use the equation of motion
S(t)= (So)+(Vo)t-(1/2)gt^2
S(20)= (So) + 140 - 240.1
(So) = -100.1 m (I know it is supposed to be a positive number)
But doesn't that give you the total height? How do I get rid of the height of the parabola above the cliff?
You have the right formula.
Let S0 be the height of the cliff in meters.
When the stone stops it is at 0m height.
So,
0m = S0 + (v0)t + (1/2)gt^2
t =7 s
g=-9.8m/(s^2)
0 = S0 + (20m/s)(7s) + (1/2)(-9.8m/(s^2))(49s^2)
0 = S0 + 140m - 240.1m
S0 = 100.1m
Assuming that from the bottom to the top of the cliff is h1 and the top of cliff to the top of the parabola is h2... isn't the 100.1m really h1 + h2 (since the stone is thrown upwards)? I am looking for just h1.
In the equation
s(t) = (s0) + (v0)t + (1/2)(a)(t^2)
s(t) gives the height for some time t.
s0 is the initial height, which is the height of the cliff.
v0 is the initial velocity, for this problem 20m/s. The last term is the distance due to gravity. As t^2 increases, this term finally overwhelms the others and the stone reverses direction and falls back to a point at the base of the cliff.
How high the stone travels above the cliff can be calculated, but it is not needed to find the height of the cliff.
To find the height of the cliff, you need to subtract the height of the parabolic path above the cliff from the total height obtained using the equation of motion.
The equation of motion you mentioned, S(t) = (So) + (Vo)t - (1/2)gt^2, gives you the position (or height) of the stone at any given time (t) during its flight. Here, "S(t)" represents the position at time "t," "(So)" represents the initial position (or height) of the stone, "(Vo)" represents the initial velocity, "g" represents the acceleration due to gravity, and "(1/2)gt^2" represents the effect of gravity causing the stone to fall.
Now, to find the height of the cliff, you need to determine the height of the parabola above the cliff. In this case, you know that the stone hits the ground after 7 seconds, which means that at that moment, its position will be zero. Therefore, you can set S(t) equal to zero and solve for time "t."
0 = (So) + (Vo)t - (1/2)gt^2
Since the initial velocity of the stone is upward, you need to use a negative value for the velocity. Plugging in the values, you get:
0 = -100.1 + (-20)(7) - (1/2)(9.8)(7)^2
Solving this equation will give you the value of "t" that corresponds to the stone hitting the ground. Once you have the value of "t," you can substitute it back into the equation of motion to find the initial height of the stone (So).
Finally, to find the height of the cliff, you subtract the height of the parabola above the cliff from the total height obtained using the equation of motion.
So, the height of the cliff = Total height - Height of the parabola above the cliff.