A ball is thrown upward with an initial velocity of 13 m/s. At what time does the ball reach the high point in its flight? (Use the approximate value of g = 10 m/s2, and remember that the velocity is equal to zero at the high point.)

use the kinematic equation vf=vi+-9.8T

(-9.8 is the true acceleration due to gravity)
what we know
vf=0
vi=13
g=-9.8
t=x
then just plug and chug answer should be around 1.326 s

Why did the ball go to the party alone? Because it wanted to have some "height"-time!

Now, let's do some math (but with a touch of humor, of course). When a ball is thrown upwards, its velocity decreases until it reaches the highest point, where it becomes zero. We can use the formula:

v = u + at,

where:
v = final velocity (in this case, 0 m/s),
u = initial velocity (13 m/s),
a = acceleration due to gravity (-10 m/s^2),
t = time taken.

Substituting the values, we have:

0 = 13 + (-10)t.

Simplifying the equation, we get:

-10t = -13.

Dividing both sides by -10, we find:

t = 1.3 seconds.

So, the ball reaches the high point in its flight at approximately 1.3 seconds. Just remember, it's always good to have a good sense of "time"ping when solving physics problems!

To find the time it takes for the ball to reach the high point in its flight, we can use the kinematic equation:

v = u + at

Where:
v = final velocity (which is 0 because the ball is at rest at the high point)
u = initial velocity (13 m/s)
a = acceleration due to gravity (-10 m/s^2)
t = time

Substituting the known values into the equation, we have:

0 = 13 - 10t

Now, we can solve for t:

10t = 13

Dividing both sides by 10, we get:

t = 1.3 seconds

Therefore, the ball will reach the high point in its flight after approximately 1.3 seconds.

To find the time it takes for the ball to reach the high point in its flight, we can use the kinematic equation for vertical motion. The equation is:

v = u + gt

Where:
v = final velocity (which is 0 at the high point)
u = initial velocity (13 m/s, thrown upward)
g = acceleration due to gravity (10 m/s^2)
t = time

Now, let's rearrange the equation to solve for time:

0 = 13 + (10)t

Subtracting 13 from both sides:

-13 = (10)t

Dividing both sides by 10:

t = -13/10

Therefore, the ball reaches the high point at approximately -1.3 seconds. However, note that time cannot be negative in this scenario since we assume the ball is thrown upward from the ground and at some positive time, it reaches the high point. Thus, we discard the negative value.

Hence, the ball reaches the high point in its flight at approximately 1.3 seconds.