During a baseball game, a batter hits a
pop-up to a fielder 78 m away. The acceleration of gravity is 9.8 m/s^2.
If the ball remains in the air for 5.7 s, how high does it rise?
Answer in units of m.
Tf = 5.7/2 = 2.85 s. = Fall time.
h = Yo*t + 0.5g*Tf^2
h = 0 + 4.9*2.85^2 = 39.8 m.
To find the height the ball rises, we can use the kinematic equation for vertical motion:
h = v₀t + 0.5at²
Where:
- h is the height,
- v₀ is the initial vertical velocity (which is 0 since the ball starts from rest),
- t is the time the ball remains in the air, and
- a is the acceleration due to gravity, which is -9.8 m/s² (negative because it acts in the opposite direction of motion).
Plugging in the values:
h = 0 + 0.5(-9.8)(5.7)²
h = 0 + 0.5(-9.8)(32.49)
h = 0 + (-4.9)(32.49)
h = -159.201
The negative sign indicates that the ball falls below the initial position. However, since we are interested in the height it rises, we take the absolute value:
h = | -159.201 |
h = 159.201 m
Therefore, the ball rises to a height of approximately 159.201 meters.