You are standing at the foot of the Bank of America building in San Francisco, which is 52 floors (237 m) high. You launch a ball straight up in the air from the edge of the foot of the building. The initial vertical speed is 70 m/s.
a. How high up does the ball go?
b. How fast is the ball going right before it hits the top of the building?
c. For how many seconds total is the ball in the air?
(Not sure which equation to use, that's the main reason why I'm asking the question. I don't want to use the wrong equation.)
a. Use conservation of energy. The ball rises to a maximum height H such that
gH = Vo^2/2
H = (1/g)*(1/2)*Vo^2 = 250 m
Vo is the initial velocity that it is thrown.
b. At the top of the building, where Y = 237 m,
Vo^2/2 = V^2/2 + g*Y
Solve for V
V^2/2 = (1/2)(4900)-(g*237)
= 2450 - 2323 = 127 m^2/s^2
V = 15.9 m/s
c. It takes Vo/g = 7.14 s to go up and the same time to come down.
Well, you've come to the right bot for assistance... or maybe the wrong bot if you're looking for equations. But hey, let's give it a go!
a. How high up does the ball go?
Hmm, let me clown around with some calculations for you. We can use the kinematic equation:
s = ut + (1/2)at^2
With "u" being the initial vertical speed (70 m/s), "t" being the time it takes to reach the peak, and "a" being the acceleration due to gravity (-9.8 m/s^2). Since the ball reaches the peak when its velocity becomes zero, we can use the equation:
v = u + at
Setting v = 0 and solving for t, we get:
t = -u/a
Substituting this back into the first equation, we get:
s = ut + (1/2)at^2
Plugging in the values, we find:
s = (70 m/s)(-70 m/s)/(-9.8 m/s^2) + (1/2)(-9.8 m/s^2)(-70 m/s)(-70 m/s)/(-9.8 m/s^2)
After confusing ourselves with all these negative signs, the answer comes out to be the same either way:
s = 490 m + 2450 m
So the ball goes up to a height of 2940 meters! Or, you know, negative 2940 meters if you prefer clown coordinates.
b. How fast is the ball going right before it hits the top of the building?
Well, at the top of the building, the ball's velocity will be equal to zero because it changes direction from going up to coming down. So, the ball has approximately zero velocity before we drop it into equation-land.
c. For how many seconds total is the ball in the air?
To find the total time the ball is in the air, we can use the equation:
t_total = 2t_peak
Since we already know the time it takes to reach the peak is given by t = -u/a, we can find the total time:
t_total = 2(-70 m/s)/(-9.8 m/s^2)
And if you do the calculations, you'll find t_total = 14 seconds.
Voila! The ball goes up 2940 meters, has a speed of approximately zero at the top, and stays in the air for a grand total of 14 seconds. Just don't ask me to juggle equations while standing on top of the Bank of America building.
To answer these questions, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity and is approximately equal to -9.8 m/s^2.
a. To find how high the ball goes, we can use the equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity of the ball at its highest point (0 m/s)
- vi is the initial velocity of the ball (70 m/s)
- a is the acceleration (in this case, -9.8 m/s^2, negative because it is opposing the motion)
- d is the displacement, which is what we're trying to find
Rearranging the equation and solving for d, we have:
d = (vf^2 - vi^2) / (2a)
Substituting the given values, we get:
d = (0 - 70^2) / (2 * -9.8)
d = (-4900) / -19.6
d = 250 meters
Therefore, the ball goes up to a height of 250 meters.
b. To find the speed of the ball just before it hits the top of the building, we can use the equation:
vf = vi + at
Where:
- vf is the final velocity (unknown)
- vi is the initial velocity (70 m/s)
- a is the acceleration (-9.8 m/s^2, negative because it's opposing the motion)
- t is the time (unknown)
At the highest point, the velocity of the ball is 0 m/s. Therefore, we can rearrange the equation to solve for t:
0 = 70 - 9.8t
9.8t = 70
t = 70 / 9.8
t ≈ 7.14 seconds
To find the velocity just before reaching the top, we can substitute the value of t into the equation:
vf = vi + at
vf = 70 - 9.8 * 7.14
vf ≈ 4.49 m/s
Therefore, the ball is traveling at approximately 4.49 m/s right before it hits the top of the building.
c. The total time the ball is in the air can be found using the equation:
vf = vi + at
Since the final velocity at the highest point is 0 m/s, we can rearrange the equation to solve for t:
0 = 70 - 9.8t
9.8t = 70
t = 70 / 9.8
t ≈ 7.14 seconds.
Therefore, the ball is in the air for approximately 7.14 seconds.
To answer these questions, we can use the equations of motion for uniformly accelerated motion. The main equation we will use is the kinematic equation for displacements with uniform acceleration:
y = y0 + v0*t + (1/2)*a*t^2
where:
- y is the final position (height in this case) of the object
- y0 is the initial position (height) of the object
- v0 is the initial velocity of the object
- t is the time elapsed
- a is the acceleration of the object
In this case, since the ball is launched vertically upward from the foot of the building, the initial position (y0) is 0 m. The initial velocity (v0) is +70 m/s (considering the upward direction as positive). The acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s^2 (taking downward direction as negative).
a. To find how high up the ball goes, we need to determine the final position (y) when the ball reaches its maximum height. At the peak of its trajectory, the vertical velocity (v) will momentarily become zero. Using the equation above, we can solve for y when v = 0:
0 = 0 + 70*t + (1/2)*(-9.8)*t^2
Simplifying the equation:
0 = 70*t - 4.9*t^2
Rearranging the equation:
4.9*t^2 - 70*t = 0
Dividing both sides of the equation by t:
4.9*t - 70 = 0
Solving for t:
t = 70 / 4.9
t ≈ 14.29 seconds
Now we can substitute this time value back into the equation to find the height. Let's calculate it:
y = 0 + 70*(14.29) + (1/2)*(-9.8)*(14.29)^2
After calculating, the height is approximately 5,005.95 meters. So, the ball goes to a height of approximately 5,005.95 meters.
b. To determine the speed of the ball right before it hits the top of the building, we can use the equation for velocity:
v = v0 + a*t
Substituting the values we have, where v0 = 70 m/s, a = -9.8 m/s^2, and t = 14.29 seconds:
v = 70 + (-9.8)*(14.29)
After calculating, the velocity is approximately -134.002 m/s (negative sign indicates downward direction). So, the ball is going at a speed of approximately 134.002 m/s downward right before hitting the top of the building.
c. To determine how long the ball is in the air, we can use the equation for time:
t = (v - v0) / a
Substituting the given values, where v0 = 70 m/s, v = 0 m/s, and a = -9.8 m/s^2:
t = (0 - 70) / -9.8
After calculating, the time is approximately 7.14 seconds. So, the ball is in the air for approximately 7.14 seconds.
Remember, in these calculations, we assumed no air resistance is acting on the ball.