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 be192.7m or 833.08m far from the throwing pointcolor(brown)((Ans.(a)))
As a projectile moves in semi- circular way, its height from the ground would same for two different timescolor(orange)((Ans.(b)))
You also want the tallest building that the projectile could clear. It means the highest heightcolor(red)('H') of the projectile from the ground
we know,color(blue)(H=(v_0^2sin^2theta_0)/(2g))=92.156mcolor(red)((Ans.(c)))