You throw a baseball directly upward at time t = 0 at an initial speed of 13.3 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.

Well, let me calculate that for you, but first, let me get my trusty clown calculator. *drumroll please* Okay, here we go.

To find the maximum height reached by the baseball, we can use the kinematic equation:

h = (V0^2) / (2g)

Where:
h is the maximum height
V0 is the initial velocity
g is the acceleration due to gravity

Plugging in the values, we get:

h = (13.3^2) / (2 * 9.8)

Calculating this, we find that the maximum height reached by the baseball is approximately 9.12 meters. That's pretty high!

Now, to find the time at which the ball passes through half the maximum height, we can use another kinematic equation:

h = V0t - (1/2)gt^2

Since we want to find when the height is half the maximum height, we can rewrite the equation as:

h/2 = (V0 * t) - (1/2) * g * t^2

Plugging in the values, we have:

(9.12/2) = (13.3 * t) - (1/2) * 9.8 * t^2

Simplifying and rearranging the equation, we get:

4.56 = 13.3t - 4.9t^2

Now, solving this quadratic equation might require a clown's touch, so I'll spare you the details and give you the two times when the ball passes through half the maximum height. The ball will reach half the maximum height at approximately t = 0.438 seconds and t = 2.42 seconds.

So, there you have it! The maximum height reached by the baseball is approximately 9.12 meters, and it will pass through half the maximum height at approximately 0.438 seconds and 2.42 seconds. Enjoy your baseball journey!

To find the maximum height the ball reaches above where it leaves your hand, we can use the kinematic equation:

h = v^2 / (2g)

where:
h is the maximum height,
v is the initial velocity of the ball (13.3 m/s),
and g is the acceleration due to gravity (9.80 m/s^2).

Plugging in the given values:

h = (13.3 m/s)^2 / (2 * 9.80 m/s^2)
= 175.89 m / 19.60
= 8.97 m

Therefore, the maximum height the ball reaches above where it leaves your hand is approximately 8.97 meters.

To find the times when the ball passes through half the maximum height, we need to consider the motion when the ball returns to the same height as when it was thrown initially (which is half the maximum height).

Using the equation:

h = h0 + v0t - (1/2)gt^2

where:
h is the height,
h0 is the initial height (0 in this case),
v0 is the initial velocity (13.3 m/s),
g is the acceleration due to gravity (9.80 m/s^2),
and t is the time.

Plugging in the values:

0 = 0 + (13.3 m/s)t - (1/2)(9.80 m/s^2)t^2

Rearranging the equation:

4.9t^2 - 13.3t = 0

Factoring out t:

t(4.9t - 13.3) = 0

Therefore, either t = 0 or 4.9t - 13.3 = 0.

Solving the second equation:

4.9t = 13.3
t = 13.3 / 4.9
t ≈ 2.71 s

Thus, the ball passes through half the maximum height at approximately 2.71 seconds after it is thrown.

To find the maximum height the ball reaches, we can use the kinematic equation for vertical motion:

h = (v^2 - u^2) / (2g)

Where:
h is the maximum height
v is the final velocity at the highest point (which is 0 since the ball stops momentarily before falling back down)
u is the initial velocity (13.3 m/s)
g is the acceleration due to gravity (9.80 m/s^2)

Substituting the given values into the equation:

h = (0^2 - 13.3^2) / (2 * 9.80)
h = (-176.89) / 19.6
h ≈ -9.02 meters

Since it doesn't make sense to have a negative height in this case, we can conclude that the maximum height the ball reaches above where it leaves your hand is approximately 9.02 meters.

To determine at what times the ball passes through half the maximum height, we can use the kinematic equation for vertical displacement:

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

Where:
d is the displacement from the starting point
u is the initial velocity (13.3 m/s)
g is the acceleration due to gravity (-9.80 m/s^2, since we are considering the downward direction)
t is the time

We need to find the value of t when d is equal to half the maximum height, which is 4.51 meters (half of 9.02 meters).

Setting d = 4.51, we can rearrange the equation:

4.51 = 13.3t + (1/2)(-9.80)t^2

Simplifying the equation:

-4.90t^2 + 13.3t + 4.51 = 0

Now we can solve this quadratic equation for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -4.90, b = 13.3, and c = 4.51.

t = (-(13.3) ± √((13.3)^2 - 4(-4.90)(4.51))) / (2(-4.90))

Simplifying further:

t = (-13.3 ± √(176.89 + 89.32)) / (-9.80)
t = (-13.3 ± √266.21) / (-9.80)

Calculating the square root:

t = (-13.3 ± 16.32) / (-9.80)

Now we can calculate two possible values for t:

t1 = (-13.3 + 16.32) / (-9.80)
t1 ≈ 0.31 seconds

t2 = (-13.3 - 16.32) / (-9.80)
t2 ≈ -3.23 seconds

Since time cannot be negative in this context, we disregard t2.

Therefore, the ball passes through half the maximum height at approximately 0.31 seconds after being thrown.

v = Vi - g t

at top v = 0
0 = 13.3 - 9.81 t
t = 1.36 seconds to top
h = Hi + Vi t - 4.9 t^2
h = 0 + 13.3 * 1.36 - 4.9 (1.36)^2
h = 18.1 - 9.06
h = 9.03 meters

when is it at 4.52 meters?
4.52 = 13.3 t - 4.9 t^2
t^2 - 2.71 t + .922 = 0

t = [ 2.71 +/- sqrt ( 7.34-3.69) ]/2
t = [ 2.71 +/- 1.91 ]/2
= .4 seconds on the way up
and
2.31 seconds on the way dwon