The initial velocity given in the question is,#v_0=125ms^(-1)#
The throwing angle, #theta_0=20°#
The mentioned height from ground,#y=58.6 m#
We know that,
#color(green)(y=v_0sintheta_0t-1/2g(t)^2#
#:.56.8=125sin20°t-1/2 9.8(t)^2#
#or,56.8=42.75t-4.9(t)^2#
#or,4.9(t)^2-42.75t+56.8=0#
#:.t=(-(-42.75)+-sqrt(42.75^2-4xx4.9xx56.8))/(2xx4.9)#
#=(42.75+-26.73)/9.8=1.64sor7.09s#
#When,t=1.64s#,then the distance from the throwing point to the hitting point of the projectile,
#x=v_0costheta_0t=125ms^(-1)xxcos20°xx1.64s=192.7m#
#When,t=7.09s#,then the distance from the throwing point to the hitting point of the projectile,#x=833.08m.#
So,The building will be#192.7m or 833.08m # far from the throwing point#color(brown)((Ans.(a)))#
As a projectile moves in semi- circular way, its height from the ground would same for two different times#color(orange)((Ans.(b)))#
You also want the tallest building that the projectile could clear. It means the highest height#color(red)('H')# of the projectile from the ground
we know,#color(blue)(H=(v_0^2sin^2theta_0)/(2g))##=92.156mcolor(red)((Ans.(c)))#
