A stone A is thrown vertically upward at 10m/s after 1 second another stone b is thrown vertically upward at 12m/s
At what height above the ground will the second stones be of the same elevation
We can start the problem by using the equation for the height h(t) of a stone thrown vertically upward, which is given by:
h(t) = h0 + v0t - (1/2)gt^2
where h0 is the initial height, v0 is the initial velocity, g is the acceleration due to gravity (g ≈ 9.8 m/s^2), and t is the time elapsed since the stone was thrown.
For stone A, we have:
hA(t) = hA(0) + vA(0)t - (1/2)gt^2
= 0 + 10t - (1/2)gt^2
= 10t - 4.9t^2 (substituting g ≈ 9.8 m/s^2)
For stone B, we have:
hB(t) = hB(0) + vB(0)t - (1/2)gt^2
= 0 + 12(t-1) - (1/2)gt^2 (since stone B is thrown 1 second later than stone A)
= 12t - 12 - 4.9t^2 (substituting g ≈ 9.8 m/s^2)
To find the height at which the two stones are at the same elevation, we need to solve the equation:
hA(t) = hB(t)
which gives:
10t - 4.9t^2 = 12t - 12 - 4.9t^2
Simplifying and rearranging, we get:
t = 1.22 seconds
Substituting this value of t into either hA(t) or hB(t), we can find the height above the ground at which the two stones are at the same elevation:
h = hA(1.22) = hB(1.22)
= 10(1.22) - 4.9(1.22)^2
= 6.10 meters
Therefore, the two stones will be at the same height of 6.10 meters above the ground after approximately 1.22 seconds.