A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 7.99 s. What is its initial velocity? Neglect air resistance.

v(t) = v(0) + ½ a t

Given:
g = -9.8m/s
t = 7.99s

Half way up the ball will reach 0 velocity.
0 = v + ¼ g t

Alternatively, all the way back, the ball will have a velocity equal but opposite to its initial.
-v = v + ½ g t

Use either equation to solve for v.

Correction:

use v(t) = v(0) + a t
v(½ t) = 0, so: 0 = v + ½ g t
v(t) = -v, so: -v = v + g t

ijiij

Well, well, well, Mr. Gravity is up to his usual antics, I see! Let me put on my detective hat and tinker with some physics equations.

When a ball is thrown vertically upward and then returns to its point of release, we know that its displacement is zero. So, the poor thing went on a little adventure just to end up where it started. Classic!

Now, let me tell you a little secret – time is a sneaky little devil. It loves to make things complicated. In this case, we have two parts to consider: the time it takes for the ball to reach its maximum height and the time it takes to fall back down.

Since the total time in the air is 7.99 seconds, we can conclude that each of these parts takes half of that time, which is approximately 3.995 seconds. Why? Because time likes to split things right down the middle.

Now, when the ball reaches its maximum height, its velocity is zero. It just hangs in the air, taking a breather. So, we only need to focus on the part of the journey when it's coming back down to us.

In this downward part, we can use one of my favorite kinematic equations:

v = u + a * t

Where:
v is the final velocity (which, in this case, is negative because gravity likes to pull things down),
u is the initial velocity (that's what we're looking for – sneaky sneaky),
a is the acceleration due to gravity (which is approximately -9.8 m/s^2, since Mr. Gravity loves negative numbers),
and t is the time it takes for the ball to fall.

Now, if we pop in the values, we have:

-9.8 m/s^2 * 3.995 s = u + (-9.8 m/s^2) * 3.995 s

Simplifying this equation, we end up with:

-39.1 m/s = u - 39.1 m/s

Since Mr. Gravity made both terms negative, they cancel each other out, leaving us with:

u = 0

So, my dear friend, the initial velocity of the ball is a grand total of zero! Which means it has no initial push to escape the earth's gravity. I guess it didn't get quite the boost it needed to party in the air for too long. Better luck next time, ball!

To find the initial velocity of the ball, we can use the following kinematic equation:

v = u + at,

where:
v = final velocity (when the ball returns to its point of release, its velocity is 0 m/s)
u = initial velocity (what we are trying to find)
a = acceleration (in this case, acceleration due to gravity is -9.8 m/s^2)
t = time taken for the ball to reach its maximum height and then return back to its point of release (which is given as 7.99 s)

Since the acceleration due to gravity is acting in the opposite direction to the positive direction, it is negative.

Using the equation v = u + at, we can substitute the given values:

0 = u + (-9.8)(7.99).

Simplifying the equation:

0 = u - 78.002.

Rearranging the equation to solve for u:

u = 78.002 m/s.

Therefore, the initial velocity of the ball is 78.002 m/s upwards.

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