Answer:
Part a)
[tex]h = 62.2 ft[/tex]
Part b)
It will not able to cross the pole
Explanation:
As we know that ball is hit with speed of 110 ft/s at an angle of 35 degree
so here we will say
[tex]v_y = 110 sin35[/tex]
[tex]v_x = 110 cos35[/tex]
now at the maximum height the vertical velocity will become zero
so here we can use kinematics
[tex]v_f^2 - v_y^2 = 2 a h[/tex]
here we have
[tex]a = -32 ft/s^2[/tex]
[tex]v_f = 0[/tex]
[tex]v_y = 63.1 ft/s[/tex]
now we have
[tex]0 - 63.1^2 = 2(-32)h[/tex]
[tex]h = 62.2 ft[/tex]
Part b)
now the height of ball is related to the distance from point of projection is given as
[tex]y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
now we know that
[tex]x = 375 ft[/tex]
[tex]y = 375(tan35) - \frac{(32)(375^2)}{2(110^2)(cos35)^2}[/tex]
[tex]y = 262.6 - 277.12 = -14.5 ft[/tex]
since its coming negative so it will not able to cross the pole