During a baseball game, a batter hits a pop- up to a fielder 78 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 5.8 s, how high does it rise?
it was at the top at time 5.8/2=2.9sec
how high was it? It fell in 2.9s
h= 1/2 g t^2=1/2 g 2.9^2
It spends 2.9s going up and 2.9s coming down.
Coming down, it falls
height = (g/2)t^2 = (4.9)*2.9^2 = 41.2 meters.
The distance of the fielder is not needed.
To find the height the ball reaches, we can use the equation of motion for vertically thrown objects. The equation is:
h = v0*t + (1/2)*a*t^2
Where:
- h is the height,
- v0 is the initial vertical velocity,
- t is the time,
- a is the acceleration.
In this case, we are given the following information:
- The ball is hit straight up in the air, so the initial vertical velocity (v0) will be positive.
- The acceleration (a) is equal to the acceleration due to gravity, which is -9.8 m/s^2 (negative because it acts downward).
- The time (t) is 5.8 seconds.
We need to find the height (h) that the ball reaches. Rearranging the equation, we have:
h = v0*t + (1/2)*a*t^2
Since the ball is hit straight up, the initial vertical velocity (v0) will be 0 m/s because there is no initial vertical speed. Substituting these values into the equation, we get:
h = 0*5.8 + (1/2)*(-9.8)*(5.8)^2
Simplifying further:
h = (-4.9)*(5.8)^2
Now we can calculate the height:
h = -4.9 * 33.64
h = -164.63 m
Since height cannot be negative, the answer is 164.63 meters.
Therefore, the ball rises approximately 164.63 meters.