During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 5.93 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat. The
acceleration of gravity is 9.8 m/s
2. Answer in units of m.

43.01 m

Explanation:

Given that :

Time ball uses in the air = 5.93, time it takes to reach maximum height and return = time of flight(T) = 5.93 s

TIME of flight (T) = 2 * time taken(t)

Where g = a = acceleration due to gravity = 9.8m/s²

S = 0.5at²

S = maximum height

Tjme taken (t) = time of flight / 2 = 5.93/2 = 2.965 s

Hence,

S = 0.5at²

S = 2 × 0.5 × 9.8 × 2.965²

S = 43.0770025

S = 43.01 m

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

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