A 47.7-g golf ball is driven from the tee with an initial speed of 44.2 m/s and rises to a height of 25.9 m. (a) Neglect air resistance and

Question

A 47.7-g golf ball is driven from the tee with an initial speed of 44.2 m/s and rises to a height of 25.9 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 7.73 m below its highest point

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Ladonna 3 months 2021-08-06T10:22:36+00:00 1 Answers 288 views 0

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    2021-08-06T10:24:23+00:00

    Answer:

    (a) 34 J

    (b) 40 m/s.

    Explanation:

    (a)As no air resistance, so total mechanical energy of golf ball remains constant.

    i.e,

    E_f=E_i

    \frac{1}{2}m(v_f)^2+mgh_f= \frac{1}{2}m(v_i)^2+mgh_i

    Here m is the mass of the golf ball and g is the acceleration due to gravity.

    Given m = 47.0 g= 0.047 kg , v_i=44.2 m/s, h_f=25.9 m and h_i=0 (because ball start from zero.)

    Substitute the given values, we get

    K.E_f+0.047kg\times9.8m/s^2\times25.9m=\frac{1}{2}\times 0.047kg\times(44.2m/s)^2 +0.047kg\times9.8m/s^2\times0

    K.E_f=33.98J=34J

    Thus, the kinetic energy of the ball at its highest point is 34 J.

    (b) when the ball is 7.73 m below its highest point then it final height,

    25.9 m – 7.73 m =18.17 m.

    Now from above,

    \frac{1}{2}m(v_f)^2+mgh_f= \frac{1}{2}m(v_i)^2+mgh_i

    \frac{1}{2}m(v_f)^2+mgh_f= \frac{1}{2}m(v_i)^2+mg\times 0

    v_f=\sqrt{v_i-2gh_f}

    Substitute the values, we get

    v_f=\sqrt{(44.2m/s)^2-2\times9.8m/s^2\times18.17m}

    v_f=39.96 m/s=40m/s

    Thus. final speed of ball when it is 7.73 m below its highest point is 40 m/s.

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