During a baseball game, a batter hits a popup
to a fielder 80 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 5.5 s, how
high does it rise?
Answer in units of m.
I do not care how far it went to the fielder.
I am only interested in the vertical problem.
it spent 5.5/2 seconds rising
v = Vi - g t
at top, v = 0
0 = Vi - 9.8 (5.5/2)
so
Vi = 27 m/s
so if you throw up at 27 m/s, how high?
two ways:
energy:
(1/2) m Vi^2 = m g h
so
h = Vi^2/(2g) = 27^2/(2*9.8) = 37 meters
or using integration
h = Vi t - 4.9 t^2
h = 27 (2.75) - 4.9 (2.75)^2
= 74 - 37
= 37 meters again
To find the height the ball rises, we can use the equations of motion in physics. In this case, we have the initial velocity (vertical velocity when the ball is hit) as 0 m/s because the ball is going straight up.
We are given the time of flight (Time the ball takes to reach the peak height) as 5.5 seconds. We can use this information to find the final velocity (vertical velocity at the peak height) and then use it again to calculate the height.
Using the equation:
vf = vo + at
Where:
vf = final velocity
vo = initial velocity
a = acceleration due to gravity
t = time
Since the initial velocity is 0 m/s, the equation becomes:
vf = 0 + 9.8 * 5.5
vf = 9.8 * 5.5
vf = 53.9 m/s
Now, we have the final velocity. To find the height, we can use the equation:
h = (vf^2 - vo^2) / (2 * a)
Where:
h = height
vf = final velocity
vo = initial velocity
a = acceleration due to gravity
Since the initial velocity is 0 m/s, the equation becomes:
h = (53.9^2 - 0^2) / (2 * 9.8)
h = (2905.21) / 19.6
h = 148.17 meters
Therefore, the ball rises to a height of approximately 148.17 meters.