A) The maximum height is 31.9 m
B) The speed of the ball is 17.7 m/s
Explanation:
A)
We can solve this problem by using the law of conservation of energy. In fact, in abcense of air resistance, the mechanical energy of the ball (sum of potential energy + kinetic energy) must be conserved.
Mathematically:
[tex]U_i +K_i = U_f + K_f[/tex]
where :
[tex]U_i[/tex] is the initial potential energy, at the bottom
[tex]K_i[/tex] is the initial kinetic energy, at the bottom
[tex]U_f[/tex] is the final potential energy, at the top
[tex]K_f[/tex] is the final kinetic energy, at the top
We can rewrite it as
[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]
where:
m is the mass of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]h_i = 0[/tex] is the initial height  of the ball
u = 25 m/s is its initial speed
[tex]h_f[/tex] is the maximum height reached by the ball
v = 0 is the final speed (which is zero at the maximum height)
Solving for [tex]h_f[/tex], we find: the maximum height:
[tex]\frac{1}{2}mu^2 = mgh_f\\h_f = \frac{u^2}{2g}=\frac{(25)^2}{2(9.8)}=31.9 m[/tex]
B)
When the ball is halfway up to its maximum height, it means that its height is
[tex]h_f = \frac{31.9}{2}=15.9 m[/tex]
Therefore we can re-apply again the equation of the conservation of energy:
[tex]mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2[/tex]
where this time v is not zero, but it is the speed of the ball at the height of 15.9 m
Re-arranging the equation and solving for v, we find:
[tex]v=\sqrt{u^2-2gh_f}=\sqrt{25^2-2(9.8)(15.9)}=17.7 m/s[/tex]
Learn more about kinetic energy and potential energy:
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