A stone is thrown vertically upward at a speed of 27.70 m/s at time t=0. A second stone is thrown upward with the same speed 2.140 seconds later. At what time are the two stones at the same height?
At what height do the two stones pass each other?
What is the downward speed of the first stone as they pass each other?
Thank you!!!
To solve this problem, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity and is approximately -9.8 m/s^2 (negative because it acts downward).
Let's start by finding the time at which the two stones are at the same height.
For the first stone:
Initial velocity (u1) = 27.70 m/s (upward)
Time (t1) = ?
Acceleration (a) = -9.8 m/s^2
Using the formula:
s1 = u1 * t1 + (1/2) * a * t1^2
where s1 is the height of the first stone at time t1.
Since the stone is thrown vertically upward, its final position (s1) will be zero when it reaches the highest point of its trajectory. Using this information, we can rewrite the equation as:
0 = u1 * t1 + (1/2) * a * t1^2
This is a quadratic equation which we can solve to find t1.
Rewriting the equation:
(1/2) * a * t1^2 + u1 * t1 = 0
Simplifying the equation:
-4.9 * t1^2 + 27.70 * t1 = 0
Factoring out t1:
t1 * (-4.9 * t1 + 27.70) = 0
Setting each factor equal to zero and solving for t1:
t1 = 0 (this is the initial time when the stone was thrown)
-4.9 * t1 + 27.70 = 0
-4.9 * t1 = -27.70
t1 = 27.70 / 4.9
t1 ≈ 5.65 seconds
Therefore, the time at which the first stone is at the same height as the second stone is approximately t1 = 5.65 seconds.
Next, let's find the height at which the two stones pass each other.
Using the equation of motion for the second stone:
Initial velocity (u2) = 27.70 m/s (upward, same as the first stone)
Time (t2) = t1 - 2.140 seconds (the second stone is thrown later)
Acceleration (a) = -9.8 m/s^2
Using the same equation for height (s) as before:
s2 = u2 * t2 + (1/2) * a * t2^2
Since both stones are at the same height when they pass each other, we can equate s1 and s2.
u1 * t1 + (1/2) * a * t1^2 = u2 * t2 + (1/2) * a * t2^2
Substituting the known values:
27.70 * 5.65 + (1/2) * (-9.8) * (5.65)^2 = 27.70 * (5.65 - 2.140) + (1/2) * (-9.8) * (5.65 - 2.140)^2
Simplifying the equation:
156.1155 ≈ 96.9013
Therefore, the height at which the two stones pass each other is approximately 96.9013 meters.
Finally, let's find the downward speed of the first stone as they pass each other.
The downward speed can be calculated as the final velocity (v1) of the first stone when it passes the height of the second stone. To find v1, we can use the equation:
v1 = u1 + a * t1
Substituting the known values:
v1 = 27.70 + (-9.8) * 5.65
Simplifying the equation:
v1 ≈ -24.02 m/s
Therefore, the downward speed of the first stone as they pass each other is approximately 24.02 m/s.
At the top point v=0 =>
time of upward motion of the 1st stone is
t=v/g =22.7/9.8=2.32 s.
Δt =2.32 -2.14 =0.18 s.
At t=2.32 s the 1st stone is at the top point
h₁=v₀t-gt²/2 =27.7•2.32 – 9.8•2.32²/2 =37.9 m
with v=0 and begins its free fall
The 2nd stone was at the point
h₂= v₀Δt-g(Δt)²/2 =27.7•0.18 – 9.8•0.18²/2 =4.83 m
and moved with upward velocity
v₂=v₀-gΔt=
=27.7-9.8•0.18=25.9m/s
Δh=37.9-4.83=33.07 m
h₁=gt²/2
h₂=v₂t- gt²/2
Δh= h₁+h₂=gt²/2+ v₂t- gt²/2 =v₂t
t= Δh/ v₂=33.07/25.9=1.28 s
h₁=gt²/2 =9.8•1.28²/2=8.03 m
The height two stones pass each other is
h₀= 37.9-8.03 =29.9 m
The time of the 1st stone motion is
2.32+1.28 = 3.6 s.
The downward speed of the first stone is
v=gt=9.8•1.28 =12.54 m/s.