During a baseball game, a batter hits a popup to a fielder 88 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 5.9 s, how
high does it rise?
Answer in units of m
To find out how high the ball rises during its trajectory, we can use the equations of motion. The key equation that helps us solve this problem is the kinematic equation:
y = v₀t + (1/2)gt²
Where:
- y is the vertical displacement or the height the ball rises (what we're trying to find)
- v₀ is the initial vertical velocity (which is zero in this case because the ball starts from rest)
- t is the time of flight of the ball (given as 5.9 seconds)
- g is the acceleration due to gravity (given as 9.8 m/s²)
Using this equation, we can substitute the given values:
y = 0 + (1/2)(9.8)(5.9)²
y = 0 + (1/2)(9.8)(34.81)
y = 0 + 169.843
y ≈ 169.84 m
Therefore, the ball rises approximately 169.84 meters during its trajectory.