Question

A stone A is thrown vertically upward at 10m/sec. Another stone b is thrown vertically upward at 12m/sec after 1 second at an initial point of B is 2 m higher than the initial point of A.
At what height that two stones have the same elevation?

Answer 1

We can start by using the kinematic equation for the height of a vertically thrown object:

h = vi*t - 0.5*g*t^2

where h is the height, vi is the initial velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s^2).

For stone A, we have:

hA = 10t - 0.5*9.8*t^2

For stone B, we have:

hB = 12(t-1) - 0.5*9.8*(t-1)^2 + 2

Note that we subtract 1 second from the time for stone B since it is thrown after 1 second, and we add 2 meters to account for the initial height difference between A and B.

To find the height at which the two stones have the same elevation, we need to find the value of t that makes hA = hB:

10t - 0.5*9.8*t^2 = 12(t-1) - 0.5*9.8*(t-1)^2 + 2

Simplifying and rearranging, we get:

4.9t^2 - 14t + 8 = 0

Using the quadratic formula, we get:

t = (14 ± sqrt(196 - 4*4.9*8)) / (2*4.9)

t ≈ 2.7 seconds or t ≈ 0.6 seconds

Note that we discard the negative solution since time cannot be negative. Therefore, the two stones have the same elevation at around t = 2.7 seconds.

To find the height at this point, we can plug t = 2.7 seconds into either hA or hB:

hA = 10*2.7 - 0.5*9.8*(2.7)^2 ≈ 37.4 meters

hB = 12(2.7-1) - 0.5*9.8*(2.7-1)^2 + 2 ≈ 37.4 meters

Therefore, the two stones have the same elevation at a height of approximately 37.4 meters.