Standing on the top ledge of a 55 m high building you throw a ball straight up with an initial speed of 45 m/s. How long to the nearest second, does it take to hit the ground?
I got 10 seconds for the first question.
How do I go about answering this one?
If the building in the previous problem is 48 meters high and you throw the ball up at 30 m/s, how high to the nearest meter does it go?
well, once it stops at the top, it is a simple fall from there.
How high above the building does it go and how long to the top?
(45 m/s is quite a throw straight up)
m g h = (1/2) m v^2
h = (1/2) (45)^2/9.81 = 103 m
average speed up = 45/2 = 22.5 m/s
so time in air up =103/22.5 = 4.58 s upward
now it falls from 55+103 = 158 m
d = (1/2) g t^2
158 = 4.9 t^2
t = 5.68 s
total time in air = 4.58 + 5.68 = 10.3 s
the way I did it the second question is really easy
The correct answer was 94 meters for the second part of the question. That's what I'm unsure as to how to get it...
To answer the first part of your question, determining the time it takes for the ball to hit the ground, we can use the kinematic equation for vertical motion:
h = ut + (1/2)gt²
where:
h = height (final position), which is 0 for hitting the ground
u = initial velocity (upward), which is 45 m/s
g = acceleration due to gravity, which is approximately 9.8 m/s²
t = time
Plugging in the values into the equation and solving for t:
0 = 45t + (1/2)(9.8)(t²)
Rearranging the equation and simplifying:
4.9t² + 45t = 0
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where a = 4.9, b = 45, and c = 0, we can substitute the values and solve for t:
t = (-45 ± √(45² - 4(4.9)(0))) / (2(4.9))
Calculating further:
t = (-45 ± √(2025)) / (9.8)
t = (-45 ± 45) / (9.8)
t = 0 or t ≈ -9.183
Since time cannot be negative in this context, we consider the positive value:
t ≈ 0 or t ≈ 9.183
Rounding to the nearest second, it takes approximately 9 seconds for the ball to hit the ground.
Now, moving on to the second part of your question, we will determine how high the ball goes if the building is 48 meters high and the initial velocity is 30 m/s.
Using the same kinematic equation as before, but this time solving for h:
h = ut + (1/2)gt²
Substituting the given values:
48 = (30)t + (1/2)(9.8)(t²)
Rearranging the equation and simplifying:
4.9t² + 30t - 48 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where a = 4.9, b = 30, and c = -48:
t = (-30 ± √(30² - 4(4.9)(-48))) / (2(4.9))
Calculating further:
t = (-30 ± √(900 + 940.8)) / 9.8
t = (-30 ± √(1840.8)) / 9.8
Since we are interested in the positive value of t, we can remove the negative sign:
t ≈ (6.985) / 9.8
t ≈ 0.713
Therefore, to the nearest meter, the ball goes approximately 1 meter high.
To find the time it takes for the ball to hit the ground, you can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
Where:
- h is the final height (ground level),
- u is the initial speed of the ball,
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time it takes for the ball to hit the ground.
In this case, the initial speed of the ball is 45 m/s, and the height is 55 meters. Since the ball is thrown straight up, its final height will be 0. Therefore, we can rearrange the equation to solve for time:
0 = 45t + (1/2)(9.8)t^2
Simplifying the equation:
4.9t^2 + 45t = 0
Now, we can solve this quadratic equation to find the time it takes for the ball to hit the ground. Plugging the equation into a solver or using the quadratic formula, we find two solutions:
t = 0 seconds (ignored, as negative time doesn't make sense in this context)
t ≈ 9.14 seconds
Since we're asked to find the time to the nearest second, the ball takes approximately 9 seconds to hit the ground.
Now, let's move on to the second question:
To find how high the ball goes when thrown up with an initial speed of 30 m/s from a building that is 48 meters high, we can use the same kinematic equation:
h = ut + (1/2)gt^2
Where:
- h is the maximum height reached by the ball,
- u is the initial speed of the ball,
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the initial speed of the ball is 30 m/s, and the height is 48 meters. We need to solve for h. We know that at the highest point, the ball's velocity is 0, so we can use this information to find the time it takes for the ball to reach its maximum height:
0 = 30 - 9.8t
Solving for t:
t ≈ 3.06 seconds
Now, we can plug this value of time back into the original equation and solve for h:
h = (30 * 3.06) + (1/2)(9.8)(3.06)^2
h ≈ 45.39 meters
Therefore, the ball reaches a height of approximately 45 meters.