A stone is dropped from the roof of a building; 1.60s after that, a second stone is thrown straight down with an initial speed of 26.0m/s , and the two stones land at the same time.
part a. How long did it take the first stone to reach the ground?
part b.How high is the building?
part c.What are the speeds of the two stones just before they hit the ground?
_Part A_
Let the first stone travel distance x(t) in and velocity u(t) in time t.
Let the second stone travel distance y(t) in and velocity v(t) in time t.
Let the second stone be released at time ŧ after the first.
Given:
x(0) = y(0) = 0[m]
u(0)=0[m/s]
v(ŧ)=26.0[m/s]
ŧ = 1.60[s]
g=9.8[m/s²]
(NB: Measuring distance down from roof.)
Find:
t such that x(t)=y(t)
Using:
x(t) = u(0)t + ½gt²
y(t) = v(ŧ)(t-ŧ) + ½g(t-ŧ)²
So using the given values and x(t)=y(t):
4.9 t² = 26.0×(t-1.60) + 4.9(t-1.60)²
Solve for t.
_Part B_
Use:
x(t) = ½gt²
_Part C_
Use:
u(t) = gt
v(t) = v(ŧ) + g(t-ŧ)
To solve this problem, we can use the equations of motion in the vertical direction.
Part a: Calculate the time it took for the first stone to reach the ground.
The equation of motion for an object in free fall is:
y = v₀t + (1/2)gt²
Given:
Initial velocity, v₀ = 0 (since the stone is dropped, its initial velocity is zero)
Time, t = 1.60 s
Acceleration due to gravity, g = 9.8 m/s²
Substituting the values into the equation, we get:
0 = 0 + (1/2)(9.8)(t₁)²
Simplifying, we have:
4.9t₁² = 0
Since the term on the left side of the equation is zero, the only solution is t₁ = 0. Therefore, it took the first stone 0 seconds to reach the ground.
Part b: Calculate the height of the building.
The equation for displacement of a falling object is:
y = v₀t + (1/2)gt²
Given:
Initial velocity, v₀ = 0 (since the stone is dropped, its initial velocity is zero)
Time, t = 1.60 s
Acceleration due to gravity, g = 9.8 m/s²
Substituting the values into the equation, we have:
y = 0 + (1/2)(9.8)(1.60)²
Simplifying, we find:
y = 0 + 12.32 m
Therefore, the height of the building is 12.32 meters.
Part c: Calculate the speeds of the two stones just before they hit the ground.
The final velocity of an object in free fall can be calculated using the equation:
v = v₀ + gt
For the first stone, since it was dropped, its initial velocity is zero. Therefore:
v₁ = 0 + 9.8(1.60)
Simplifying, we find:
v₁ = 15.68 m/s
For the second stone, the initial velocity is given as 26.0 m/s. Therefore:
v₂ = 26.0 + 9.8(1.60)
Simplifying, we find:
v₂ = 41.68 m/s
Therefore, the speed of the first stone just before it hits the ground is 15.68 m/s, and the speed of the second stone is 41.68 m/s.
To solve this problem, we need to use the equations of motion. Let's break down each part of the question and solve them step by step.
Part a: How long did it take the first stone to reach the ground?
The first stone is dropped from rest, which means it has an initial velocity of 0 m/s. We can use the equation for free fall motion to find the time it takes for the stone to reach the ground:
h = (1/2) * g * t^2
Where:
h = height of the building
g = acceleration due to gravity (9.8 m/s²)
t = time taken for the stone to reach the ground
Since the stone is dropped from rest, the initial height is equal to the height of the building. Rearranging the equation, we can solve for t:
t = sqrt(2h / g)
Part b: How high is the building?
In this part, we need to find the height of the building. We can use the time it took the first stone to reach the ground, which we found in part a, and substitute it into the equation of motion to find the height:
h = (1/2) * g * t^2
Part c: What are the speeds of the two stones just before they hit the ground?
To find the speed of an object just before it hits the ground, we can use the equation:
v = u + g * t
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (9.8 m/s²)
t = time taken for the stone to reach the ground
Since the second stone is thrown with an initial speed of 26.0 m/s, we can substitute the values into the equation to find the final velocity.
Now, let's solve each part step by step.
Part a: How long did it take the first stone to reach the ground?
- Use the equation t = sqrt(2h / g) and plug in the values given to find the time it took for the first stone to reach the ground.
Part b: How high is the building?
- Use the equation h = (1/2) * g * t^2 and plug in the time from part a to find the height of the building.
Part c: What are the speeds of the two stones just before they hit the ground?
- Use the equation v = u + g * t and plug in the values given for the second stone to find its final velocity. The final velocity of the first stone will be 0 m/s as it is dropped from rest.
I hope this explanation helps you understand how to solve the problem.