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

Answer 1

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.