a bot throws a cricket ball witg an initial velocity of 5m.s from the edge of a tall building.it passes the edge of the building on its way down and strikes the ground below after 5seconds (neglect effects of friction. question1...how does the acceleration of the ball moving downwards compare with the upwards movement?calculate the time taken react the higher point.calculate the height the building.draw a position vs time graph inserting all known values.

V^2 = Vo^2 + 2g*h

h = (V^2-Vo^2)/2g
h = (0-25)/-19.6 = 1.28 m. = Max. ht.
above bldg.

V = Vo + g*t
Tr = (V-Vo)/g = (0-5)/-9.8 = 0.51 s. =
Rise time.

Tf1 = Tr = 0.51 s. = Time to fall back to edge of bldg.

Tf2 = 5 = - Tr - Tf1=5-0.51 - 0.51=3.98 s.

Hb = Vo*t + 0.5a*t^2
Hb = 5*3.98 + 4.9*(3.98)^2 = 97.5 m. =
ht. of bldg.

T = Time in seconds.

h = Ht above gnd. in meters.

(T,h).
(0,97.5)
(0.51,98.8)
(1.01,97.5)
(5,0).

To analyze the given scenario, we can break it down into different parts and answer each question step by step.

Question 1: How does the acceleration of the ball moving downwards compare with the upwards movement?

Acceleration can be calculated using the equation:

acceleration = change in velocity ÷ time

When the ball is moving upwards, it experiences a negative acceleration (deceleration) due to the effect of gravity, which is equal to the acceleration due to gravity, denoted as "g" and approximately equal to 9.8 m/s².

When the ball is moving downwards, it experiences a positive acceleration due to the force of gravity acting in the same direction as the motion. Therefore, the acceleration of the ball moving downwards is equal to the acceleration due to gravity, which is 9.8 m/s².

So, the acceleration of the ball moving downwards and upwards is the same, both equal to 9.8 m/s².

Next, let's calculate the time taken to reach the higher point.

Since we know the initial velocity of the ball (5 m/s) and the acceleration (9.8 m/s²), we can use the equation of motion:

final velocity = initial velocity + (acceleration × time)

When the ball reaches the highest point, its final velocity becomes zero. So, we can rearrange the equation as follows:

0 = 5 m/s + (9.8 m/s² × time)

Solving for time:

9.8 m/s² × time = -5 m/s
time = -5 m/s ÷ 9.8 m/s²

Now, since time cannot be negative, we take the positive value of time:

time = 0.51 s (approximately)

Therefore, it takes approximately 0.51 seconds for the ball to reach the higher point.

Now, let's calculate the height of the building.

Using the equation of motion again:

final velocity² = (initial velocity)² + 2 × acceleration × displacement

When the ball reaches the ground, its final velocity is 0, and the initial velocity is 5 m/s. The acceleration is -9.8 m/s² (negative because it is in the opposite direction of the motion).

0 = (5 m/s)² + 2 × (-9.8 m/s²) × displacement

Solving for displacement:

(5 m/s)² = 2 × 9.8 m/s² × displacement
25 m²/s² = 19.6 m/s² × displacement
displacement = 25 m²/s² ÷ (19.6 m/s²)

displacement ≈ 1.28 m

Therefore, the height of the building is approximately 1.28 meters.

Finally, let's draw a position vs. time graph.

In the given scenario, the position of the ball changes over time. Initially, the ball starts from the top of the building with a position of 0 and gradually increases its distance from the building's edge as it moves downwards. We can assume the positive direction of motion as downwards.

On the position vs. time graph, the x-axis represents time (in seconds), and the y-axis represents position (in meters). We can plot the following points:

(0, 0) - starting point
(5, 1.28) - point where the ball strikes the ground

Connecting these two points with a line will give us the position vs. time graph.

I hope this explanation helps! Let me know if you have any further questions.