A stone thrown from the top of a building is given an initial velocity of 40.0 m/s straight upward. The building is 70.0m high, and the stone just misses the edge of the roof on its way down, determine
(A) The time at which the stone reaches its maximum height,
(B) The maximum height,
(C) The time at which the stone returns to the height from which it was thrown,
(D) The velocity of the stone at this instant
(E) The velocity and position of the stone at t = 10.00 s.
The key to this is to come up with the equation of motion. If you review the relevant section, you will easily see that
h(t) = 70+40t-4.9t^2
v(t) = 40-9.8t
Now just apply your skills from Algebra I to find the vertex and find h(t)=70 and v(10)
To solve this problem, we can use the equations of motion for an object in free fall. There are four main equations we can use:
1) v = u + at
2) s = ut + (1/2)at^2
3) v^2 = u^2 + 2as
4) s = vt - (1/2)gt^2
where:
v is the final velocity
u is the initial velocity
a is the acceleration (which is -9.8 m/s^2 for objects in free fall near the Earth's surface)
t is the time
s is the displacement or position
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's solve the problem step by step:
(A) The time at which the stone reaches its maximum height:
To find this, we need to find the time it takes for the stone to reach its maximum height. At this point, the velocity will be zero.
Using the first equation:
v = u + at
Since the final velocity is zero, we can write the equation as:
0 = 40 - 9.8t
Simplifying the equation:
9.8t = 40
Finding t:
t = 40 / 9.8
(B) The maximum height:
Now that we have the time it takes for the stone to reach its maximum height, we can find the maximum height using the second equation:
s = ut + (1/2)at^2
Using the given values:
s = 40t + (1/2)(-9.8)(t^2)
Substituting the value of t we found in part (A):
s = 40(40 / 9.8) + (1/2)(-9.8)((40 / 9.8)^2)
Simplifying the equation will give the maximum height.
(C) The time at which the stone returns to the height from which it was thrown:
To find this, we need to find the time it takes for the stone to reach the height of the building, which is 70 meters.
Using the fourth equation:
s = vt - (1/2)gt^2
Substituting the given values:
70 = vt - (1/2)(9.8)(t^2)
Solving this equation will give the time at which the stone returns to the height from which it was thrown.
(D) The velocity of the stone at this instant:
To find the velocity of the stone at this instant, we can use the first equation:
v = u + at
Substituting the values:
v = 40 + (-9.8)t
Calculating the value of v at the time found in part (C) will give the velocity at that instant.
(E) The velocity and position of the stone at t = 10.00 s:
Using the equations of motion, we can find the velocity and position of the stone at this time. Simply substitute the value of t into the equations and calculate the results.
By following these steps and using the equations of motion, you will be able to answer all the parts of the problem.
To solve this problem, we need to use the kinematic equations of motion.
Let's start with part A. The time at which the stone reaches its maximum height.
The first step is to find the time it takes for the stone to reach its highest point (maximum height). We can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the stone is thrown straight upward, the initial velocity is positive 40.0 m/s, and the acceleration due to gravity is negative 9.8 m/s^2.
At the highest point, the final velocity will be zero. Plugging the values into the equation:
0 = 40.0 - 9.8t,
Simplifying it:
9.8t = 40.0,
t = 40.0 / 9.8,
t ≈ 4.08 s.
So, the time at which the stone reaches its maximum height is approximately 4.08 seconds.
Moving on to part B, the maximum height.
We can use the following equation:
s = ut + 0.5at^2,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
At the highest point, the displacement will be equal to the maximum height reached. Plugging in the values:
s = 40.0(4.08) + 0.5(-9.8)(4.08)^2,
Simplifying it:
s ≈ 83.84 m.
So, the maximum height reached by the stone is approximately 83.84 meters.
Now, let's move on to part C, the time at which the stone returns to the height from which it was thrown.
Since we know the time it took to reach the maximum height, we can double that to find the total time it takes for the stone to return to its original height.
So, the time at which the stone returns to the height from which it was thrown is approximately 2 * 4.08 = 8.16 seconds.
Next, part D, the velocity of the stone at the instant it reaches its initial height.
Using the equation:
v = u + at,
we can plug in the values:
v = 40.0 - 9.8(8.16),
v ≈ -40.0 m/s.
So, the velocity of the stone when it reaches its initial height is approximately -40.0 m/s. The negative sign indicates that the stone is moving downward.
Finally, part E, the velocity and position of the stone at t = 10.00 s.
The velocity can be found using the same equation as before:
v = u + at,
v = 40.0 - 9.8(10.0),
v ≈ -58.0 m/s.
So, the velocity of the stone at t = 10.00 s is approximately -58.0 m/s.
To find the position of the stone at t = 10.00 s, we can use the equation:
s = ut + 0.5at^2.
Plugging in the values:
s = 40.0(10.0) + 0.5(-9.8)(10.0)^2,
s ≈ 190.0 m.
So, the position of the stone at t = 10.00 s is approximately 190.0 meters.