Question

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

Answer 1

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.