During a baseball game, a batter hits a popup to a fielder 81 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 6.7 s, how
high does it rise?
It rises the same distance that an object will fall in 6.7/2 = 3.35 seconds, because that is the time that it will take the ball to fall from maximum height.
H = (1/2)*g*(3.35)^2
To determine the height the ball rises, we can use the kinematic equation that relates the distance, time, initial velocity, and acceleration.
The equation we will use is:
distance = initial velocity * time + 0.5 * acceleration * time^2
Given:
- Initial distance = 81 m
- Time = 6.7 s
- Acceleration = 9.8 m/s^2 (acceleration due to gravity)
- Initial velocity is not given, but since the ball is hit directly upwards, we can assume it starts with an initial velocity of 0 m/s.
Substituting the values into the equation, we get:
81 m = 0 m/s * 6.7 s + 0.5 * 9.8 m/s^2 * (6.7 s)^2
Simplifying the equation:
81 m = 0 + 0.5 * 9.8 m/s^2 * (6.7^2 s^2)
We can calculate this mathematically to find the height the ball rises.