during a baseball game, a batter hits a popup to a fielder 64 m away. the acceleration of gravity is 9.8 m/s 2 . if the ball remains in the air for 4.4 s, how high does it rise? answer in units of m.



Answer :

The maximum height of the ball during parabolic motion is 94.864 m.

We need to know about the parabolic motion to solve this problem. Parabolic motion is an object's motion under acceleration with an initial angle of elevation. It should follow the rule

vx = vo . cosA

vy = vo . sinA

tmax = vy / g

ymax = vy² / 2g

where vx is horizontal velocity, vy is vertical velocity, vo is initial velocity, A is angle of the projectile, g is gravitational acceleration, tmax is maximum interval time and ymax is maximum height.

From the question above, the given parameters are

g = 9.8 m/s²

xmax = 64 m

tmax = 4.4 s

Find the relationship between maximum height and maximum time

ymax = vy² / 2g

ymax = tmax² . g / 2

ymax = 4.4² . 9.8 / 2

ymax = 94.864 m

Find more on the parabolic motion at: https://brainly.com/question/1259873

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