A ball is thrown 233 m upward and then falls back to earth.
Neglecting air resistance, how long will it be in the air? The acceleration of gravity is 9.8 m/s2 . Answer in units of s.
use a free fall equation ... h = -1/2 g t^2 + 233
set h equal to zero and solve for t
... t is the time to fall from 233 m , so double it for the full flight time
Well, this is going to be a real "up and down" adventure for the ball! Let's see how long our little friend stays in the air.
When the ball is thrown upward, it will experience a negative acceleration due to gravity. The acceleration due to gravity is constant at 9.8 m/s², and in this case, it acts in the opposite direction to the ball's initial velocity.
To find the time the ball spends in the air, we can use the fact that the final velocity of the ball at the highest point is 0 m/s. We can use the following equation to calculate the time:
Vf = Vi + at
Where:
Vf is the final velocity (0 m/s),
Vi is the initial velocity (which we'll assume is positive, as it's thrown upward),
a is the acceleration (-9.8 m/s²), and
t is the time we're trying to find.
Plugging in the values:
0 m/s = Vi - 9.8 m/s² × t
We can rearrange the equation to solve for time:
t = Vi / 9.8 m/s²
Since the initial velocity is positive (as it's thrown upward), let's plug in the value of Vi = 233 m/s:
t = 233 m/s / 9.8 m/s²
By dividing these numbers, we get:
t ≈ 23.78 s
So, our little ball will spend approximately 23.78 seconds in the air. Just long enough to enjoy the view before coming back down to Earth!
To find the time it takes for the ball to be in the air, we can use the equations of motion.
First, we need to find the time it takes for the ball to reach its highest point (when it reaches its maximum height). The initial velocity at this point is zero, and the acceleration due to gravity is acting in the opposite direction. We can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Since the ball is thrown upwards, its initial velocity is positive, and the acceleration is negative due to gravity. So, the equation becomes:
0 = vi - gt
Solving for t, we get:
t = vi / g
where g is the acceleration due to gravity. Plugging in the values, we have:
t = 233 m / 9.8 m/s^2
t ≈ 23.78 s
So, it takes approximately 23.78 seconds for the ball to reach its highest point.
Next, to find the total time in the air, we need to consider the time it takes for the ball to fall back down to the ground. This is the same as the time it takes to reach the highest point, since the ball follows the same path when it falls back down.
Therefore, the total time in the air is:
2t = 2 * 23.78 s
= 47.56 s
So, neglecting air resistance, the ball will be in the air for approximately 47.56 seconds.
To find the time it takes for the ball to be in the air, we can use the equation of motion:
h = v₀t + (1/2)gt²
where:
- h is the height reached by the ball (233 m),
- v₀ is the initial velocity (which we assume to be zero as the ball is thrown upwards),
- t is the time in the air, and
- g is the acceleration due to gravity (-9.8 m/s², taking direction into account).
Since the ball reaches a maximum height (233 m) and then falls back to the ground, we can divide the time into two parts: the time it takes to reach the highest point and the time it takes to fall back to the ground.
First, let's find the time it takes to reach the highest point. At the highest point, the velocity will be zero. So, using the equation of motion:
0 = v₀ + gt
Since v₀ is zero when the ball is thrown upward, the equation simplifies to:
t₁ = -v₀ / g
Now, let's find the time it takes for the ball to fall back to the ground. Since the initial velocity at this point is also zero, we can use the equation:
h = (1/2)gt²
Substituting the values:
233 m = (1/2)(-9.8 m/s²)t²
Simplifying and rearranging the equation:
t² = (-2 × 233 m) / -9.8 m/s²
t² = 47.3469 s²
Taking the square root of both sides:
t = √47.3469 s
t ≈ 6.883 s
Now, to find the total time in the air, we add the time taken to reach the highest point (t₁) and the time taken to fall back to the ground (t):
Total time = t₁ + t
= -v₀ / g + √(2h / g)
= 0 + √(2 × 233 m / 9.8 m/s²)
≈ √(476 / 9.8) s
≈ √(48.5714) s
≈ 6.972 s
Therefore, neglecting air resistance, the ball will be in the air for approximately 6.972 seconds.