A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can re

Question

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

in progress 0
niczorrrr 3 years 2021-09-04T16:05:39+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-04T16:07:16+00:00

    Answer:

    h = 16.9 m

    Explanation:

    When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

    Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

    (0.5)m(Vf² – Vi²) = mgh

    h = (0.5)(Vf² – Vi²)/g

    where,

    Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

    Vi = Initial Speed of Ball = 18.2 m/s

    g = acceleration due to gravity = – 9.8 m/s² ( negative for upward motion)

    h = maximum height the ball can reach = ?

    Therefore, using values in the equation, we get:

    h = (0.5)[(0 m/s)² – (18.2 m/s)²]/(-9.8 m/s²)

    h = 16.9 m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )