A ball is thrown upward. What is its time interval between the release of he ball and the time it reaches its maximum height? Its initial vertical speed is 9 m/s and the acceleration of gravity is9.8 m/s^. Neglect air resistance.
Answer in units of s.
Well, well, well, we have a ball soaring through the skies! How exciting! Now, let me calculate the time interval for you, my inquisitive friend.
When a ball is thrown upward, it reaches its maximum height when its vertical velocity becomes zero before turning around and making its descent. In this case, the initial vertical speed is 9 m/s, and gravity's acceleration is 9.8 m/s².
Now, let's play with some numbers. We know that the initial velocity is 9 m/s, and the acceleration due to gravity is -9.8 m/s² (negative because gravity is pulling the ball downwards). We also know that at the maximum height, the velocity is zero.
Using the equation (final velocity) = (initial velocity) + (acceleration * time), and plugging in the values:
0 = 9 - 9.8 * t,
where 't' is the time interval we are trying to find.
Now, if we rearrange the equation and solve for 't', we get:
t = 9 / 9.8 ≈ 0.918 s.
There you have it! The time interval between the release of the ball and when it reaches its maximum height is approximately 0.918 seconds.
But hey, don't worry, the fun doesn't end there! The ball will come back down, and we can have more adventures with gravity and time. Wheee!
To find the time interval between the release of the ball and when it reaches its maximum height, we can use the following equation for vertical motion:
v = u + at
Where:
v = final velocity (which is 0 m/s at the maximum height)
u = initial velocity (9 m/s, directed upward)
a = acceleration due to gravity (-9.8 m/s², directed downward)
t = time
Plugging in the values, we have:
0 = 9 - 9.8t
Rearranging the equation to solve for t:
9.8t = 9
t = 9 / 9.8
Calculating this, we find:
t ≈ 0.918 s
Therefore, the time interval between the release of the ball and when it reaches its maximum height is approximately 0.918 seconds.
To find the time interval between the release of the ball and the time it reaches its maximum height, we can use the kinematic equation that relates displacement, initial velocity, final velocity, and time. The equation is:
v_f = v_i + a * t
Where:
v_f is the final velocity (which is zero when the ball reaches its maximum height),
v_i is the initial velocity,
a is the acceleration (in this case, the acceleration due to gravity, which is -9.8 m/s^2 because it acts downwards),
and t is the time interval we want to find.
Given that the initial velocity is 9 m/s and the acceleration due to gravity is 9.8 m/s^2, we can plug these values into the equation:
0 = 9 + (-9.8) * t
Simplifying the equation:
-9.8 * t = -9
Dividing both sides by -9.8:
t = -9 / -9.8
t = 0.91837 s
Rounding to the appropriate number of significant figures, the time interval between the release of the ball and its maximum height is approximately 0.92 seconds.
V = Vo + g*t = 0 @ max ht.
t = -Vo/g = -9/-9.8 = 0.92 s.