Here in this question as we can see there is no air friction so we can use the principle of energy conservation
[tex]PE_i + KE_i = PE_f + KE_f[/tex]
[tex]mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2[/tex]
now here we know that
[tex]h_1 = 2.75 m[/tex]
[tex]v_i = 0[/tex]
[tex]v_f = 5.23 m/s[/tex]
now plug in all values in above equation
[tex]mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2[/tex]
divide whole equation by mass "m"
[tex]9.8*2.75 = 9.8*h + \frac{1}{2}*27.35[/tex]
[tex]9.8*h = 13.27[/tex]
[tex]h = 1.35 m[/tex]
so height of the ball from ground will be 1.35 m
