A stone is dropped off a cliff of height h. At the same time, a second stone is thrown straight upward from the base of the cliff with an initial velocity . Assuming that the second rock is thrown hard enough, at what time t will the two stones meet?

Answer 1

To solve this problem, we need to find the time it takes for each stone to reach the same height.

Let's denote the time it takes for the first stone (dropped off the cliff) to reach its peak as t1, and the time it takes for the second stone (thrown upward) to reach the same height as t2.

The first stone experiences free fall, so we can use the equation:

h = (1/2)gt1^2

Where h is the height of the cliff and g is the acceleration due to gravity.

The second stone experiences uniform acceleration in the opposite direction of gravity, so we can use the equation:

0 = v0t2 - (1/2)gt2^2

Where v0 is the initial velocity of the second stone (thrown upward).

Since the two stones meet at the same height, we can equate the two equations:

(1/2)gt1^2 = v0t2 - (1/2)gt2^2

Rearranging the equation, we get:

(1/2)gt1^2 + (1/2)gt2^2 = v0t2

Combining like terms:

(1/2)gt1^2 + (1/2)gt2^2 - v0t2 = 0

Multiplying through by 2 to get rid of the fractions:

gt1^2 + gt2^2 - 2v0t2 = 0

Now we can solve for t2 using the quadratic formula:

t2 = (-b ± √(b^2 - 4ac))/(2a)

Where a = g/2, b = -2v0, and c = g(t1^2).

Plugging in the values:

t2 = (-(-2v0) ± √((-2v0)^2 - 4(g/2)(g(t1^2)))) / (2(g/2))

Simplifying:

t2 = (2v0 ± √(4v0^2 - 2*g^2*t1^2)) / g

Simplifying further:

t2 = (2v0 ± √(4(v0^2 - g^2t1^2))) / g

We want the positive value because time cannot be negative:

t2 = (2v0 + √(4(v0^2 - g^2t1^2))) / g

Now we have the time it takes for the second stone to reach the same height as the first stone.