Explanation:
Given
height of building (h)=54 m
Projectile velocity=330 m/s
initial vertical velocity[tex](u_y)=0[/tex]
Initial horizontal velocity[tex](u_x)=330 m/s[/tex]
Time taken to cover vertical height of 54 m
[tex]h=ut+\frac{gt^2}{2}[/tex]
[tex]54=0+\frac{9.81\cdot t^2}{2}[/tex]
[tex]t=\sqrt{\frac{54\times 2}{9.81}}[/tex]
t=3.31 s
Horizontal distance traveled in this time
[tex]R_x=u_x\times t[/tex]
[tex]R_x=330\times 3.31=1094.94 m[/tex]
Vertical component of velocity when it hits the ground
v=u+at
[tex]v=0+9.81\times 3.31=32.47 m/s[/tex]
