v = 7.65 - 9.81 t
h = 11.1 + Vi t - 4.9 t^2
0 = 11.1 + 7.65 t - 4.9 t^2
solve quadratic for t, time at h = 0, ground, use the positive answer :)
then do
v = 7.65 - 9.81 t for the speed
h = 11.1 + Vi t - 4.9 t^2
0 = 11.1 + 7.65 t - 4.9 t^2
solve quadratic for t, time at h = 0, ground, use the positive answer :)
then do
v = 7.65 - 9.81 t for the speed
2.01 for time
Step 1: Determine the time it takes for the stone to reach its maximum height:
Using the vertical motion equation:
h = ut + (1/2)gt^2
Where:
h = height (11.1 m)
u = initial velocity (7.65 m/s)
g = acceleration due to gravity (-9.81 m/s^2)
t = time
Let's rearrange the equation:
11.1 = (7.65)t - (1/2)(9.81)t^2
Step 2: Solve the quadratic equation for t:
Rearrange the equation to:
(1/2)(9.81)t^2 - (7.65)t + 11.1 = 0
Solving this quadratic equation, we get two values for t: t1 = 0.399 s and t2 = 2.212 s.
Since it takes 2.212 seconds for the stone to reach maximum height and fall back to its starting point, we can conclude that it will take twice this time for the stone to hit the ground.
So, the total time for the stone to be in the air is 2 × 2.212 = 4.424 seconds.
Step 3: Determine the speed of the stone when it impacts the ground:
Using the equation for horizontal motion:
v = u + gt
Where:
v = final velocity (what we need to find)
u = initial velocity (7.65 m/s)
g = acceleration due to gravity (-9.81 m/s^2)
t = total time in the air (4.424 s)
v = 7.65 - 9.81 × 4.424
v ≈ -39.75 m/s (rounded to two decimal places)
Since speed is always positive, the speed of the stone when it impacts the ground is approximately 39.75 m/s.
1. Stone going up:
When the stone is released, its initial velocity is 7.65 m/s (upward) and it is moving against gravity. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Here, the final velocity will be 0 m/s (since the stone reaches its highest point and momentarily stops).
The acceleration (a) is the acceleration due to gravity, which is -9.81 m/s^2 (negative because it acts in the opposite direction).
The displacement (s) is the vertical distance traveled in the upward motion, which is equal to the initial height of the building, 11.1 m.
Using the equation v^2 = u^2 + 2as and plugging in the values:
0 = (7.65)^2 + 2(-9.81)(11.1)
0 = 58.8225 - 216.942
-158.1195 = -216.942
The negative sign indicates that the stone doesn't go up in this case. The stone only falls down.
2. Stone falling down:
Now, we can use the same equation v^2 = u^2 + 2as to find the final velocity (v), but with u as -7.65 m/s (negative because it is in the downward direction) and a as -9.81 m/s^2. The displacement (s) will be the distance from the initial point of release to the ground, which is 11.1 m.
v^2 = (-7.65)^2 + 2(-9.81)(11.1)
v^2 = 58.8225 + (-2)(-9.81)(11.1) [Applying the negative sign for acceleration]
v^2 = 58.8225 + 216.942
v^2 = 275.7645
v = √275.7645
v ≈ 16.61 m/s
Therefore, the speed at which the stone impacts the ground is approximately 16.61 m/s.
To find the time the stone is in the air, we can use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity v will be -16.61 m/s (negative because it is in the downward direction), the initial velocity u is -7.65 m/s, and the acceleration a is -9.81 m/s^2.
Using the equation v = u + at and plugging in the values:
-16.61 = -7.65 + (-9.81)t
-16.61 + 7.65 = -9.81t
-8.96 = -9.81t
t = -8.96 / -9.81
t ≈ 0.91 s
Therefore, the stone is in the air for approximately 0.91 seconds.