During a baseball game, a batter hits a high pop- up. If the ball remains in the air for 6.0s, how high does it rise? Hint: calculate the height using the second half of the trajectory

maybe this help?

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 calculate the height that the ball rises during its trajectory, you can use the equation of motion for projectile motion. The formula for the vertical displacement of an object in projectile motion is given by:

d = (V0 * t) - (0.5 * g * t^2)

where:
d = vertical displacement or height
V0 = initial vertical velocity (i.e., the velocity at the beginning of the trajectory)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

In this case, we can use the second half of the trajectory (when the ball is rising) to calculate the height. Since the ball remains in the air for 6.0 seconds, we divide this time by 2 to get the time it takes for the ball to reach its maximum height:

t = 6.0s / 2 = 3.0s

Now, we can plug these values into the equation:

d = (V0 * t) - (0.5 * g * t^2)

Since the ball is at maximum height at this point, its vertical velocity (V0) will be 0, as it momentarily stops and changes direction. Therefore, we can simplify the equation to:

d = - (0.5 * g * t^2)

Now, let's substitute the actual values:

d = - (0.5 * 9.8 m/s^2 * (3.0s)^2)

Calculating this expression will give us the height to which the ball rises during its trajectory.

176.4m

The hint makes it two easy. From what height does it require for something to fall in three seconds.

h=1/2 g t^2