To solve this problem, we need to find the time t at which the height of the first stone dropped off the cliff is equal to the height of the second stone thrown upward from the base of the cliff.
Let's denote the height of the cliff as h and the initial velocity of the second stone as v.
For the first stone dropped off the cliff, we can use the formula for free fall to find its height at time t:
h1(t) = h - (1/2)gt^2
For the second stone thrown upward, we can use the formula for motion with constant acceleration to find its height at time t:
h2(t) = vt - (1/2)gt^2
We want to find the time t when h1(t) is equal to h2(t):
h - (1/2)gt^2 = vt - (1/2)gt^2
Simplifying the equation, we get:
h = vt
Dividing both sides of the equation by v, we get:
t = h/v
Therefore, the two stones will meet at time t = h/v.
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 v �� i . Assuming that the second rock is thrown hard enough, at what time t will the two stones meet?
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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 v �� i . Assuming that the second rock is thrown hard enough, at what time t will the two stones meet?
To solve this problem, we can find the time it takes for the second stone to reach a height of h.
The height of the first stone dropped off the cliff at time t is given by the equation:
h1(t) = h - (1/2)gt^2, where g is the acceleration due to gravity.
The height of the second stone thrown upward at time t is given by the equation:
h2(t) = v0t - (1/2)gt^2, where v0 is the initial velocity of the second stone.
Since we want to find the time when the two stones meet, we set h1(t) = h2(t) and solve for t:
h - (1/2)gt^2 = v0t - (1/2)gt^2
Simplifying the equation, we get:
h = v0t
Dividing both sides of the equation by v0, we get:
t = h/v0
Therefore, the two stones will meet at time t = h/v0.
The height of the first stone dropped off the cliff at time t is given by the equation:
h1(t) = h - (1/2)gt^2, where g is the acceleration due to gravity.
The height of the second stone thrown upward at time t is given by the equation:
h2(t) = v0t - (1/2)gt^2, where v0 is the initial velocity of the second stone.
Since we want to find the time when the two stones meet, we set h1(t) = h2(t) and solve for t:
h - (1/2)gt^2 = v0t - (1/2)gt^2
Simplifying the equation, we get:
h = v0t
Dividing both sides of the equation by v0, we get:
t = h/v0
Therefore, the two stones will meet at time t = h/v0.
1 answer