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 w

Question

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.

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Đan Thu 4 years 2021-07-18T03:13:27+00:00 1 Answers 17 views 0

Answers ( )

  1. minhkhang
    0
    2021-07-18T03:15:21+00:00
    Reply

    Answer: 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

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