A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak.

To determine the height to which the ball rises before it reaches its peak, we can use the equation for vertical displacement:

d = v₀t + 0.5at²

where:
d is the vertical displacement (height),
v₀ is the initial vertical velocity,
t is the time,
a is the acceleration due to gravity.

In this case, the ball is popped straight up, so its initial vertical velocity is 0 (v₀ = 0). The acceleration due to gravity is directed downwards and has a value of -9.8 m/s².

So, the equation simplifies to:

d = 0.5 * (-9.8) * t²

Now we can plug in the given value for time:

d = 0.5 * (-9.8) * (6.25)²

Solving this equation gives us the height to which the ball rises before it reaches its peak.

To determine the height to which the ball rises before it reaches its peak, we can use the equations of motion for vertical motion. The key equation for this problem is:

h = v₀t + (1/2)gt²

where:
h is the height
v₀ is the initial vertical velocity
t is the time
g is the acceleration due to gravity

In this case, since the baseball is popped straight up, the initial vertical velocity (v₀) is 0 m/s because it is thrown vertically upward. The acceleration due to gravity (g) is -9.8 m/s² because it acts downward.

Substituting these values into the equation, we have:

h = 0(6.25) + (1/2)(-9.8)(6.25)²

Simplifying the equation:

h = (1/2)(-9.8)(6.25)²
h = (-4.9)(6.25)²
h = (-4.9)(39.06)
h = -191.094

Since the height cannot be negative, we take the absolute value to get the positive value:

h ≈ 191.094

Therefore, the height to which the ball rises before it reaches its peak is approximately 191.094 meters.

v=u at

0=u (-9.8) (25/8)
u=30.5m/s

s=ut 1/2 (at�0…5)
s=30.5(25/8) 1/2(-9.8)(25/8)�0…5
s=(1525/16) (-3062.5/64)
s=1525/16 - 30625/640
61000-30625/640
30375/640
47.46m