At t = 0, a stone is dropped from a cliff above a lake; 2.0 seconds later another stone is thrown downward from the same point with an initial speed of 33 m/s. Both stones hit the water at the same instant. Find the height of the cliff.
I have just recently had to answer a very similar question. My prof. has given me the two general eqns to use:
Y(final)=Y(initial)+V(initial)((T(final)-T(initial))+1/2a(T(final)-T(initial))^2
and
V(final)=V(initial)+a(T(final)-T(initial))
From there, write down your knowns and derive more specific eqns from the general ones... Very vague i know...good luck
Y1 = -(g/2) t^2
Y2 = -33(t-2) -(g/2)((t-2)^2
Set Y1 - Y2 and solve for t.
Use that t to compute Y at that time.
when i solve for t I get an extremely small number like .1998. I don't think I am setting up the problem right or something.
To find the height of the cliff, we can use the equations of motion and the concept of free fall.
Let's break down the problem into two parts: the first stone dropped from the cliff and the second stone thrown downward.
1. Stone Dropped from the Cliff:
The time it takes for the first stone to hit the water is 2.0 seconds longer than when it was released. Therefore, the time of flight for the first stone is the time it takes for the second stone to hit the water, which is t.
Using the equation of motion for vertical motion:
h = (1/2) * g * t^2
Where:
h is the height of the cliff
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time of flight for the second stone (which is also the time it takes for the first stone to hit the water).
2. Stone Thrown Downward:
The second stone is thrown downward with an initial speed of 33 m/s. Since it is thrown downward, the initial velocity is negative.
Let's denote the height of the cliff as h. The equation of motion for this stone can be written as:
h = -33t - (1/2) * g * t^2
Where:
h is the height of the cliff
t is the time of flight for the second stone.
Since both stones hit the water at the same time, the height h will be the same for both equations. Therefore, we can equate these two equations:
(1/2) * g * t^2 = -33t - (1/2) * g * t^2
Simplifying the equation:
g * t^2 = -66t
Dividing both sides by t:
gt = -66
Now, we know that g = 9.8 m/s^2, so we can substitute that into our equation:
9.8t = -66
Solving for t:
t = -66 / 9.8
t ≈ -6.7 s
Since time cannot be negative in this context, we can disregard the negative value and take the positive value of t ≈ 6.7 s.
Finally, we can substitute this value of t back into any of the equations to find the height of the cliff. Let's use the equation of motion for vertical motion for the first stone:
h = (1/2) * g * t^2
h = (1/2) * 9.8 * (6.7^2)
h ≈ 224.2 m
Therefore, the height of the cliff is approximately 224.2 meters.