A stone is thrown up vertically from the ground (the gravitational acceleration is g = 10 m/s^2 ). After a time dt =1 s, a second stone is thrown up vertically. The first stone has an initial speed v1 = 11.0 m/s, and the second stone v2 = 16.0 m/s.
(a) At what time after the first stone is thrown will the two stones be at the same altitude above ground? (in seconds)
a. h1 = V1*t + 0.5g*t^2
h1 = 11*1 - 5*1^2 = 6 m. Head start.
V^2 = V1^2 + 2g*h
V^2 = 11^2 - 20*6 = 1
V = 1.0 m/s.
h2=V2*t + 0.5g*t^2 = 6 + V*t + 0.5g*t^2
16*t - 5*t^2 = 6 + 1*t - 0.5g*t^2
The 0.5g*t^2 terms cancel:
16t - t = 6
15t = 6.0
t = 0.4 s. to catch up.
To find the time at which the two stones will be at the same altitude above the ground, we can use the kinematic equation for vertical motion:
h = h0 + v0t - (1/2)gt^2
where:
h is the height above the ground,
h0 is the initial height (which we can assume is zero for both stones),
v0 is the initial velocity,
t is the time, and
g is the gravitational acceleration.
Let's first find an equation for each stone:
For the first stone:
h1 = v1t - (1/2)gt^2
For the second stone:
h2 = v2t - (1/2)gt^2
Since we want to find the time at which both stones are at the same altitude, we set h1 = h2:
v1t - (1/2)gt^2 = v2t - (1/2)gt^2
Simplifying, we get:
v1t = v2t
Dividing both sides by t:
v1 = v2
Now we can solve for t:
t = v1 / v2
Substituting the given values of v1 = 11.0 m/s and v2 = 16.0 m/s into the equation:
t = 11.0 m/s / 16.0 m/s
Simplifying, we get:
t ≈ 0.6875 s
Therefore, after approximately 0.6875 seconds, the two stones will be at the same altitude above the ground.