During a baseball game, a batter hits a pop-up to a fielder 94 m away.
The acceleration of grvity is 9.8 m/s^2. If the ball remains in the air for 5.5 seconds, how high does it rise? Answer in units of m.
A projectile staying in the air for t seconds spends half the time going up and half going down.
Since it is a deceleration, we can think of it a being dropped from the highest point and stayed in the air for 2.5 seconds.
S=ut+(1/2)gt²
=0+(1/2)g*2.5²
To find the height to which the ball rises, we can use the kinematic equation for vertical motion:
y = v0t + (1/2)at^2
Where:
y is the vertical displacement or height
v0 is the initial vertical velocity (0 m/s because the ball starts at rest)
t is the time
a is the acceleration due to gravity (-9.8 m/s^2, negative because it acts downward)
In this case, we are given the time (t = 5.5 seconds) and the acceleration due to gravity (a = -9.8 m/s^2). We need to find the vertical displacement (y).
Plugging in the known values into the equation:
y = (0 m/s)(5.5 s) + (1/2)(-9.8 m/s^2)(5.5 s)^2
y = 0 + (-4.9 m/s^2)(30.25 s^2)
y = -147.725 m^2/s^2
Notice that we have meters squared per second squared, which is not a unit for displacement. This is because we used the equation for height, which includes the acceleration due to gravity. To find the height, we need to divide this value by 2.
y = (-147.725 m^2/s^2) / 2
y = -73.8625 m^2/s^2
Now, we have the correct units, but the value is negative. This shows that the ball has risen in the opposite direction, so we take the absolute value to find the magnitude of the height.
y = | -73.8625 m^2/s^2 |
y = 73.8625 m^2/s^2
Finally, we remove the squared units and express the height in meters:
y = sqrt(73.8625 m^2/s^2)
y = 8.59 m
Therefore, the ball rises to a height of approximately 8.59 meters.