During a baseball game, a batter hits a high pop-up. if the ball remains in the air for 4.2 seconds, how high does it rise? The accelleration of gravity is 9.8 m/s^2.
ignoring the height of contact
time up equals time down
h = .5 g t^2 = .5 * 9.8 * 2.1^2
idk what he wrote so here ya go
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
To determine the height the baseball rises, we can use the following formula:
h = (1/2) * g * t^2
where:
h = height the baseball rises
g = acceleration due to gravity (9.8 m/s^2)
t = time the ball remains in the air (4.2 seconds)
Substituting the values into the formula:
h = (1/2) * 9.8 m/s^2 * (4.2 s)^2
First, square the time:
h = (1/2) * 9.8 m/s^2 * 17.64 s^2
Now multiply the values:
h = 0.5 * 9.8 m/s^2 * 17.64 s^2
h = 88.2 m
Therefore, the baseball rises to a height of 88.2 meters.