A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does the ball have

Question

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does the ball have a speed of 12m/s?

in progress 0
Helga 7 months 2021-07-30T07:41:41+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-07-30T07:43:04+00:00

    Answer:

    (A) The maximum height of the ball is 40.57 m

    (B) Time spent by the ball on air is 5.76 s

    (C) at 33.23 m the speed will be 12 m/s

    Explanation:

    Given;

    initial velocity of the ball, u = 28.2 m/s

    (A) The maximum height

    At maximum height, the final velocity, v = 0

    v² = u² -2gh

    u² = 2gh

    h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m

    (B) Time spent by the ball on air

    Time of flight = Time to reach maximum height + time to hit ground.

    Time to reach maximum height = time to hit ground.

    Time to reach maximum height  is given by;

    v = u – gt

    u = gt

    t = \frac{u}{g}

    Time of flight, T = 2t

    T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s

    (C) the position of the ball at 12 m/s

    As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

    v² = u² – 2gh

    12² = 28.2² – 2(9.8)h

    12² – 28.2² = – 2(9.8)h

    -651.24 = -19.6h

    h = 651.24 / 19.6

    h = 33.23 m

    Thus, at 33.23 m the speed will be 12 m/s

Leave an answer

Browse

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