A ball is thrown vertically upward with a speed of 19.0 m/s.

(a) How high does it rise?
m

(b) How long does it take to reach its highest point?
s

(c) How long does the ball take to hit the ground after it reaches its highest point?
s

(d) What is its velocity when it returns to the level from which it started?


m/s

a. h = (Vf^2-Vo^2)/2g.

h = (0-361)/-19.6 = 18.4 m.

b. Tr = (Vf-Vo)/g.
Tr = (0-19)/-9.8 = 1.94 s. = Rise time.

c. Tf = Tr = 1.94 s. = Fall time.
OR
0 + 4.9*t^2 = 18.4.
t^2 = 3.755
Tf = 1.94 s. = Fall time.

d. Vf = Vo + gt = 0 + 4.9*1.94=19 m/s.

(a) How high does it rise?

Well, since I can't actually see the ball, I'll have to do some math for you. The formula for calculating the height is h = (v^2) / (2g), where v is the initial velocity and g is the acceleration due to gravity. So, plugging in the numbers we have: h = (19.0^2) / (2 * 9.8). Let me just calculate that for you... and the height is approximately 18.44 meters.

(b) How long does it take to reach its highest point?
To find the time it takes to reach the highest point, we can use the formula t = v / g, where v is the initial velocity and g is the acceleration due to gravity. Plugging in the numbers: t = 19.0 / 9.8. So, it takes around 1.94 seconds to reach the highest point.

(c) How long does the ball take to hit the ground after it reaches its highest point?
Since the ball falls freely after reaching its highest point, we can use the formula t = sqrt(2h / g), where h is the height and g is the acceleration due to gravity. Plugging in the numbers: t = sqrt(2 * 18.44 / 9.8). After some calculations, the ball takes approximately 2.42 seconds to hit the ground.

(d) What is its velocity when it returns to the level from which it started?
When the ball returns to its initial level, its velocity is the same magnitude as the initial velocity, but in the opposite direction. So, the velocity when it returns is -19.0 m/s. Don't worry, it's not a negative ball, just a negative velocity!

To solve this problem, we can use the kinematic equations of motion.

(a) To find the maximum height reached by the ball, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

In this case, the initial velocity is 19.0 m/s, the final velocity is 0 m/s (at the highest point, the ball momentarily comes to rest), and the acceleration due to gravity (assuming no air resistance) is -9.8 m/s^2 (negative because the acceleration is downward).

Plugging the values into the equation, we have:

0 = (19.0)^2 + 2(-9.8)s

0 = 361 - 19.6s

19.6s = 361

s = 361/19.6

s ≈ 18.47 m

Therefore, the ball rises to a height of approximately 18.47 meters.

(b) The time taken to reach the highest point can be calculated using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

In this case, the initial velocity is 19.0 m/s, the final velocity is 0 m/s, and the acceleration is -9.8 m/s^2.

Plugging the values into the equation, we have:

0 = 19.0 + (-9.8)t

9.8t = 19.0

t = 19.0/9.8

t ≈ 1.94 s

Therefore, it takes approximately 1.94 seconds for the ball to reach its highest point.

(c) The time taken to hit the ground after reaching the highest point is the same as the time taken to reach the highest point. Therefore, it is approximately 1.94 seconds.

(d) The velocity of the ball when it returns to the level from which it started can be calculated using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

In this case, the initial velocity is 19.0 m/s, the final velocity is unknown, the acceleration is -9.8 m/s^2, and the time taken is twice the time taken to reach the highest point, which is approximately 1.94 seconds.

Plugging the values into the equation, we have:

v = 19.0 + (-9.8)(2*1.94)

v ≈ 19.0 - 38.0

v ≈ -19.0 m/s

Therefore, the velocity of the ball when it returns to the level from which it started is approximately -19.0 m/s. Note that the negative sign indicates that the velocity is in the opposite direction of the initial velocity.

To solve this problem, we can use the equations of motion for freely falling objects. These equations are:

1. v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

3. v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given:
- Initial velocity (u) = 19.0 m/s

(a) To calculate the maximum height reached by the ball, we need to find the displacement (s) when the velocity (v) becomes zero.
Using equation 2, we have:
0 = (19.0) * t - (1/2) * 9.8 * t^2, where a = -9.8 m/s^2 (acceleration due to gravity)
Rearranging and solving the quadratic equation, we get two solutions for t: t = 0s (at the start) and t = 3.86s (at the highest point).
To find the maximum height, substitute the value of t into equation 2:
s = (19.0) * 3.86 - (1/2) * 9.8 * (3.86)^2

(b) To calculate the time taken to reach the highest point, we already found it in part (a): t = 3.86s.

(c) To calculate the time taken to hit the ground after reaching the highest point, we can use the same equation 2.
We now consider the downward motion with an initial velocity of 0 m/s and solve for t:
s = (0) * t - (1/2) * 9.8 * t^2
Since we are interested in the time taken to hit the ground after reaching the highest point, the displacement is equal to the maximum height. Solving the quadratic equation for t, we get two solutions: t = 0s (at the highest point) and t = 3.86s (when it hits the ground after reaching the highest point).
Therefore, the time taken to hit the ground after reaching the highest point is 3.86s.

(d) To calculate the velocity when the ball returns to the level from which it started, we note that the final velocity will have the same magnitude as the initial velocity but opposite direction.
Therefore, the velocity when it returns to the level from which it started is -19.0 m/s (or 19.0 m/s downward).