a person throws a ball upward into the air with an initial velocity of 15 m/s calculate

a)how high it goes?
b)how long the ball is in the air before it comes back ?
c)how much time it takes for the ball to reach the maximum height?

h = 15t - 4.9t^2 = 4.9t(3.06-t)

(c) max height is reached at t = 15/9.8

(b) h=0 when t=0 or 3.06

(a) calculate h using (a)

To calculate the answers to the given questions, we can use the formulas of motion under constant acceleration.

a) To find the height the ball reaches, we need to determine the maximum height it reaches during its upward motion.

The formula to find the maximum height is:
v_f^2 = v_i^2 + 2 * a * d

Where:
v_f = final velocity (0 m/s at the highest point)
v_i = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2, since it is acting against the upward motion)
d = displacement (height)

Rearranging the formula, we have:
d = (v_f^2 - v_i^2) / (2 * a)

Substituting the given values, we get:
d = (0^2 - 15^2) / (2 * -9.8)
d = 225 / -19.6
d ≈ -11.5 meters

Therefore, the ball reaches a height of approximately 11.5 meters before it starts falling back down.

b) To find the time the ball is in the air before it comes back, we need to consider both the upward and downward motions of the ball.

Using the formula:
t = (v_f - v_i) / a

For the upward motion:
v_f = 0 m/s (at the highest point)
v_i = 15 m/s
a = -9.8 m/s^2

Substituting these values, we find the time taken for the upward motion:
t_upward = (0 - 15) / (-9.8)
t_upward ≈ 1.53 seconds

For the downward motion, the initial velocity is 0 m/s at the highest point, and we use the same acceleration.

Using the formula, t = (v_f - v_i) / a, we find the time for the downward motion:
t_downward = (0 - 0) / (-9.8)
t_downward = 0 seconds

To calculate the total time the ball is in the air, we need to add both these times:
Total time = t_upward + t_downward
Total time ≈ 1.53 seconds + 0 seconds
Total time ≈ 1.53 seconds

So, the ball is in the air for approximately 1.53 seconds before it comes back down.

c) To find the time it takes for the ball to reach the maximum height, we only need to consider the upward motion.

Using the formula t = (v_f - v_i) / a, we can find the time taken:
t_upward = (0 - 15) / (-9.8)
t_upward ≈ 1.53 seconds

Therefore, it takes approximately 1.53 seconds for the ball to reach the maximum height.

To calculate the answers to the given questions, we can use the kinematic equations of motion for vertical motion.

a) To find the maximum height reached by the ball, we need to use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2, assuming standard gravitational acceleration)
s = displacement

By plugging in the given values into the equation, we can solve for s:

0^2 = 15^2 + 2(-9.8)s

225 = -19.6s

s = -225 / -19.6

Therefore, the height reached by the ball is approximately 11.48 meters.

b) To calculate the time the ball is in the air before it comes back down, we can use the equation:

v = u + at

Where:
v = final velocity (0 m/s when the ball reaches the ground)
u = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time

By plugging the given values into the equation, we can solve for t:

0 = 15 + (-9.8)t

-15 = -9.8t

t = -15 / -9.8

The time the ball is in the air before it comes back down is approximately 1.53 seconds.

c) To determine how much time it takes for the ball to reach its maximum height, we can use the equation:

v = u + at

Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (15 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time

By plugging the given values into the equation, we can solve for t:

0 = 15 + (-9.8)t

-15 = -9.8t

t = -15 / -9.8

Therefore, it takes approximately 1.53 seconds for the ball to reach its maximum height.