a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. at what time is the ball 20 m above the ground?

See your previous post: Sun,12-1-13,10:34 PM.

h = Vo*t + 0.5g*t^2 = 20

37.8t _ 4.9t^2 = 20
-4.9t^2 + 37.8t - 20 = 0
Use Quadratic Formula and get:
t = 0.571, and 7.14 s.

So, when the ball is rising; it reaches
20 m in 0.571 s. When it is falling, it
reaches 20 m in 7.14 s.

To find the time at which the baseball is 20 m above the ground, you can use the equations of motion. Specifically, you can use the kinematic equation for displacement in the vertical direction:

h = (vi × t) + (0.5 × a × t^2)

Where:
h = displacement (in this case, 20 m)
vi = initial velocity (37.8 m/s)
a = acceleration (in this case, taking g = 9.8 m/s^2 as the acceleration due to gravity)
t = time

We want to find the time when the displacement h is 20 m. To do this, we rearrange the equation:

20 = (37.8 × t) + (0.5 × 9.8 × t^2)

Simplifying this equation, we get:

4.9t^2 + 37.8t - 20 = 0

This is a quadratic equation, which we can solve using various methods. One convenient method is to use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Applying this formula to our equation, we get:

t = (-37.8 ± √(37.8^2 - 4 × 4.9 × -20)) / (2 × 4.9)

After evaluating this expression, we can find the two possible values of t. However, since time cannot be negative in this context, we discard the negative solution. Thus, we only consider the positive value of t.

Plug in the values into the equation and calculate the positive value of t:

t = (-37.8 + √(37.8^2 - 4 × 4.9 × -20)) / (2 × 4.9)

t ≈ 1.54 seconds

Therefore, when the baseball is 20 m above the ground, it will occur at approximately 1.54 seconds after it leaves the bat.