Q.1 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.

Q.II An object is thrown downward with an initial speed of 17 m/s from a height of 96 m above the ground. At the same instant, a second object is propelled vertically up from ground level with a speed of 22 m/s. At what height above the ground will the two objects pass each other? The acceleration of gravity is 9.8 m/s2 .

A.1 Well, if the ball is thrown upwards and then falls back down, it's basically having a case of gravity-induced indecisiveness. It can't make up its mind whether to go up or down. But luckily, we don't have to worry about air resistance in this scenario.

To find out how long it will be in the air, we can use the fact that what goes up must come down. The time it takes for the ball to reach its maximum height is the same as the time it takes for it to come back down.

So, we can find the time it takes for the ball to reach its maximum height using the equation:

time = (2 * initial velocity) / acceleration

Plugging in the values, we get:

time = (2 * 0) / 9.8

Wait, the initial velocity is 0 at the highest point! Silly ball, taking a break up there. So the time it takes for the ball to reach its maximum height is zero.

And since the time it takes to go up is the same as the time it takes to come back down, the total time the ball is in the air is 2 times zero, which is still zero.

In other words, the ball doesn't stay in the air at all! It goes up and comes back down in no time. It's like a really quick skydiving trip without the parachute.

So, the answer is 0 seconds.

A.II Now, let's talk about these two objects playing a game of gravity tag. One is thrown downward and the other is propelled vertically up from the ground. They're like two ships passing in the night, but instead of ships, they're objects, and instead of night, it's the daytime.

To figure out where they'll pass each other height-wise, we will calculate the time it takes for each object to reach that point, and then use that time to find the height.

The time it takes for the first object to reach the passing point can be calculated by taking the height of the first object and dividing it by its initial downward velocity:

time1 = height1 / velocity1

Similarly, the time it takes for the second object to reach the passing point can be calculated by taking the height of the second object and dividing it by its upward velocity:

time2 = height2 / velocity2

Now, let's plug in the values given:
height1 = 96 m
velocity1 = 17 m/s
height2 = ?? (the height above the ground where they meet)
velocity2 = 22 m/s

Using the equations above, we can find the respective times:
time1 = 96 / 17
time2 = height2 / 22

And since the two objects meet at the same time, we can equate the two time equations:

time1 = time2

96 / 17 = height2 / 22

Now, solving for height2:

height2 = (96 / 17) * 22

Calculating...
height2 = 128

So, the two objects will pass each other at a height of 128 meters above the ground. It's like a game of gravity peek-a-boo! But remember, no cheating by looking down or up!

Q.1 To find the time the ball will be in the air, we can use the kinematic equation:

Distance = (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)

Since the ball starts at ground level, the initial velocity is 0 m/s and the initial height is 0 m. The distance traveled by the ball is 233 m.

Plugging in the values into the equation, we get:

233 = 0.5 * (-9.8) * Time^2

Simplifying the equation, we have:

233 = -4.9 * Time^2

Dividing both sides by -4.9, we get:

Time^2 = -233 / -4.9

Time^2 = 47.55

Taking the square root of both sides, we find:

Time = √47.55

Time ≈ 6.9 seconds

Therefore, the ball will be in the air for approximately 6.9 seconds.

Q.II To find the height at which the two objects pass each other, we can calculate the time it takes for each object to reach that point.

For the first object thrown downwards, the initial velocity is 17 m/s and the initial height is 96 m. The acceleration due to gravity is -9.8 m/s^2.

Using the kinematic equation:

Final Height = Initial Height + (Initial Velocity * Time) + (0.5 * Acceleration * Time^2)

Let's assume the time it takes for the two objects to pass each other is t seconds.

For the first object thrown downwards:

Final Height = 96 + (17 * t) + (0.5 * -9.8 * t^2)

For the second object thrown upwards:

Final Height = 0 + (22 * t) + (0.5 * 9.8 * t^2)

Since the two objects pass each other, their final heights will be the same. Therefore:

96 + (17 * t) + (0.5 * -9.8 * t^2) = (22 * t) + (0.5 * 9.8 * t^2)

Simplifying the equation:

96 + 17t - 4.9t^2 = 22t + 4.9t^2

Combining like terms:

9.8t^2 + 5t - 96 = 0

Now we can solve this quadratic equation for t using any suitable method.

Using the quadratic formula:

t = (-5 ± √(5^2 - 4 * 9.8 * -96)) / (2 * 9.8)

Calculating the value inside the square root:

t = (-5 ± √(25 + 3763.2)) / 19.6

t = (-5 ± √3788.2) / 19.6

Since we are looking for a positive time, we can ignore the negative case:

t ≈ 4.07 seconds

Now, to find the height at which the two objects pass each other, we can substitute this time back into one of the original equations.

Let's use the equation for the second object:

Final Height = 0 + (22 * 4.07) + (0.5 * 9.8 * 4.07^2)

Simplifying the equation, we get:

Final Height ≈ 0 + 89.74 + 80.97

Final Height ≈ 170.71 meters

Therefore, the two objects will pass each other at a height of approximately 170.71 meters above the ground.

To answer question 1, we can use the equations of motion, specifically the one relating distance, time, initial velocity, and acceleration. In this case, the initial velocity is the velocity with which the ball is thrown upward, and the distance is the total vertical distance covered by the ball (233 m).

First, we need to find the time taken for the ball to go up to the highest point (where its velocity becomes zero) and then come back down to the ground. Since the acceleration due to gravity is acting in the opposite direction during the upward motion, we can use the equation:

v = u + at

Where:
v is the final velocity (zero in this case as the ball reaches its highest point),
u is the initial velocity (the upward velocity with which the ball is thrown),
a is the acceleration due to gravity (-9.8 m/s^2),
and t is the time taken.

Rearranging the equation to solve for time (t), we get:

t = (v - u) / a

Substituting the given values, v = 0 m/s, u = initial upward velocity, and a = -9.8 m/s^2, we can solve for time.

Now to calculate the time taken for the ball to go up and come back down, we need to find the time taken for the upward motion and multiply it by 2, as the downward motion will take the same amount of time. Once we have the time, we can specify the answer in units of seconds.

To answer question II, we can apply the concept of relative motion. We need to find the point where the two objects meet, which means both objects will have covered the same distance vertically.

First, we need to find the time it takes for the first object to reach the ground. We can use the equation of motion:

v^2 = u^2 + 2as

Where:
v is the final velocity (which is 0 m/s when the object reaches the ground),
u is the initial velocity (given as 17 m/s),
a is the acceleration due to gravity (9.8 m/s^2),
and s is the distance traveled (which is the initial height of 96 m).

Rearranging the equation to solve for time (t), we get:

t = (v - u) / a

Substituting the given values, v = 0 m/s, u = 17 m/s, and a = 9.8 m/s^2, we can solve for time.

Next, we need to find the time it takes for the second object to reach the point of intersection. We can use the same equation of motion:

t = (v - u) / a

Substituting the given values, v = 0 m/s (as the object reaches the highest point where velocity becomes zero), u = 22 m/s, and a = -9.8 m/s^2, we can solve for time.

Since the two objects are moving towards each other and have started at different heights, we can calculate the distance covered by each object using the formula:

s = ut + (1/2)at^2

Where:
s is the distance covered,
u is the initial velocity,
t is the time taken,
and a is the acceleration.

By calculating the distance covered by each object at the time of intersection, we can determine the height above the ground where the objects meet.

It will be in the air twice as long as it takes to fall from 233 meters high

h = (1/2) g t^2
233 = (1/2)(9.8) t^2
solve for t
multiply by two
==========================
height and time the same
h1 = h2
t1 = t2

h = Hi + Vi t - 4.9 t^2
so
h1 = 96 - 17 t - 4.9 t^2
h2 = 0 + 22 t - 4.9 t^2
so
96 -17 t = 22 t
96 = 39 t
t = 96/39 seconds