A ball is thrown upwards at 15m/s from a window 20 metres above the ground. Find: a)How long the ball the takes to reach the level of the window

b)How much longer it takes to reach the ground?
c)What speed it takes to reach the ground?

Correction:

c. V^2 = (15)^2 + 19.6*20 = 617
V = 24.8 m/s.

a. = V = Vo + g*t

Tr = (V-Vo)/g = (0-15)/-9.8 = 1.53 s. =
Rise time.
Tf = Tr = 1.53 s. = Fall time.
T = Tr + Tf = 1.53 + 1.53 = 3.06 s. =
Time to return to window level.

b. H = Vo*t + 0.5g*t^2 = 20 m.
15*t + 4.9t^2 = 20
4.9t^2 + 15t - 20 = 0
Use Quadratic Formula.
T = 1.0 s. To reach Gnd.

c. V^2 = Vo^2 + 2g*h.
V^2 = 15 + 19.6*20 = 392
V = 19.8 m/s.

To solve this problem, we can use the equations of motion. Let's break it down step-by-step:

a) How long does the ball take to reach the level of the window?

The initial velocity of the ball is 15 m/s in the upward direction, and it is thrown from a height of 20 meters. We can use the equation:

h = ut + (1/2)gt^2

Where:
h = height
u = initial velocity
t = time
g = acceleration due to gravity

In this case, we need to find the time taken to reach the level of the window, so let's consider the height as 20 meters. Since the ball is thrown upward, the acceleration due to gravity acts in the opposite direction, so we can take it as -9.8 m/s^2.

Using the equation, we have:

20 = (15)t + (1/2)(-9.8)t^2

Rearranging the equation:

-4.9t^2 + 15t - 20 = 0

Using the quadratic formula, we get:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we have:

t = (-15 ± √(15^2 - 4(-4.9)(-20))) / (2(-4.9))

Simplifying:

t ≈ (-15 ± √(225 - 392)) / (-9.8)

t ≈ (-15 ± √(-167)) / (-9.8)

Since the square root of a negative number is not real, it means that the ball will not reach the level of the window again. Therefore, there is no solution for part a) in this case.

b) How much longer does it take to reach the ground?

Since the ball does not reach the level of the window again, it will fall continuously towards the ground. We can find the time it takes to reach the ground using the equation:

h = ut + (1/2)gt^2

In this case, the initial height is 20 meters, and the acceleration due to gravity is -9.8 m/s^2. We need to find the time when the height (h) is 0. Thus, the equation becomes:

0 = (1/2)(-9.8)t^2

Simplifying the equation:

-4.9t^2 = 0

Dividing both sides by -4.9, we get:

t^2 = 0

Taking the square root, we have:

t = 0

Since t = 0, it means that the ball will reach the ground instantly.

c) What speed does it take to reach the ground?

To find the speed at which the ball hits the ground, we can use the equation:

v = u + gt

Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately -9.8 m/s^2)

In this case, since the ball was thrown upwards, the initial velocity is 15 m/s and the acceleration due to gravity is acting in the opposite direction (-9.8 m/s^2).

Plugging the values into the equation, we have:

v = 15 + (-9.8)(0)

Simplifying:

v = 15

Therefore, the speed at which the ball hits the ground is 15 m/s.

To solve this problem, we can make use of the kinematic equations of motion in physics. Let's break down each part of the question.

a) How long does the ball take to reach the level of the window?

We can use the kinematic equation for displacement:

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

Where:
- s is the displacement (which is 20 meters in this case)
- u is the initial velocity (15 m/s)
- t is the time taken
- a is the acceleration (which, in this case, is the acceleration due to gravity, -9.8 m/s^2, since the ball is moving upward)

Rearranging the equation, we get:

t = (v - u) / a

Substituting in the given values:

t = (0 - 15) / (-9.8)

Simplifying the equation:

t ≈ 1.53 seconds

Therefore, the ball takes approximately 1.53 seconds to reach the level of the window.

b) How much longer does it take to reach the ground?

Since the initial velocity is given as upward, we can assume the final velocity when it reaches the ground is 0 m/s. Using the kinematic equation:

v = u + at

Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (15 m/s)
- a is the acceleration (-9.8 m/s^2)

Rearranging the equation, we get:

t = (v - u) / a

Substituting in the given values:

t = (0 - 15) / (-9.8)

Simplifying the equation:

t ≈ 1.53 seconds

Therefore, it will take approximately another 1.53 seconds for the ball to reach the ground after reaching the level of the window.

c) What speed does it take to reach the ground?

To find the speed when the ball reaches the ground, we can use the equation we derived in part b:

v = u + at

Substituting in the given values:

v = 15 + (-9.8) × 1.53

Simplifying the equation:

v ≈ -0.14 m/s

Therefore, the ball reaches the ground with a speed of approximately -0.14 m/s. The negative sign indicates that the velocity is in the opposite direction of the initial velocity, which is downward.