A stone is thrown vertically upward at a speed of 27.00 m/s at time t=0. A second stone is thrown upward with the same speed 2.060 seconds later. At what time are the two stones at the same height?
To solve this problem, we can use the equations of motion for objects under the influence of gravity.
Let's assume the initial height of both stones is h = 0 (ground level). The initial velocity of both stones is 27.00 m/s, but the second stone begins 2.060 seconds later.
For the first stone, its velocity decreases due to the gravitational pull until it reaches its maximum height before falling down.
The equation for the height of the first stone as a function of time (t) can be given by:
h₁ = v₀t - (1/2)gt²
For the second stone, it starts with an initial delay of 2.060 seconds but has the same initial velocity as the first stone. So the equation for the height of the second stone as a function of time (t) can be given by:
h₂ = v₀(t - 2.060) - (1/2)gt²
Since we want to find the time at which the two stones are at the same height, we need to solve for the time (t) when h₁ = h₂.
v₀t - (1/2)gt² = v₀(t - 2.060) - (1/2)gt²
Simplifying the equation:
v₀t - (1/2)gt² = v₀t - 2.060v₀ - (1/2)gt²
We can see that the terms with t² on both sides cancel out:
-2.060v₀ = 0
This equation simplifies further to:
v₀ = 0
This equation shows that at time t = 2.060 seconds (when the second stone is thrown), both stones will be at the same height.
Therefore, the two stones are at the same height 2.060 seconds after the second stone is thrown.