During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.11 s, how high does it rise?
vf-vi=at, so find that negative number then divide by two and it will give u |vi|
make it positive
vf =0 at the top, so
0^2=vi^2 + 2a ∆x
then
-(vi^2)/(2a){should be 19.6}= ∆x
242
To determine how high the ball rises during its flight, we can use the equations of motion. First, we need to identify the known values and quantities involved:
1. The ball remains in the air for 4.11 seconds (time, t = 4.11 s).
2. The initial velocity of the ball is zero (u = 0 m/s) since it starts from rest.
3. The acceleration due to gravity is constant and equal to 9.8 m/s^2 (acceleration, a = 9.8 m/s^2).
To find the height the ball rises, we can use the following equation of motion:
s = ut + (1/2)at^2
Where:
s is the vertical displacement (height),
u is the initial velocity,
t is the time, and
a is the acceleration.
Since the initial velocity was zero, the equation simplifies to:
s = (1/2)at^2
Now, we can substitute the known values into the equation:
s = (1/2) * 9.8 m/s^2 * (4.11 s)^2
Calculating this expression will give us the height the ball rises during its flight:
s = (1/2) * 9.8 m/s^2 * 16.8921 s^2
s ≈ 84.4605 meters (rounded to four decimal places)
Therefore, the ball rises approximately 84.4605 meters during its flight.