Given data
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*The given height of the building is h = 120 m
*The given initial velocity is v_i = 72.0 m/s
*The angle above the horizontal is
[tex]\theta=25^0[/tex]*The value of acceleration due to gravity is g = 9.80 m/^2
The initial horizontal velocity of the projectile is calculated as
[tex]\begin{gathered} v_x_{}=v_i\cos \theta \\ =(72.0)\cos 25^0 \\ =65.25\text{ m/s} \end{gathered}[/tex]The initial vertical component velocity of the projectile is calculated as
[tex]\begin{gathered} v_y_{}_{}=v_i\sin \theta \\ =(72.0)\sin 25^0 \\ =30.42\text{ m/s} \end{gathered}[/tex]The time taken by the projectile is calculated by the kinematic equation of motion as
[tex]H=v_yt+\frac{1}{2}gt^2[/tex]Treating the vertically upward direction to be a negative and the vertically downward direction to be positive.
Substitute the known values in the above expression as
[tex]\begin{gathered} 120=-(30.42)t+\frac{1}{2}(9.8)t^2 \\ t=8.95\text{ s} \end{gathered}[/tex]The expression for the distance traveled by the projectile is given as
[tex]D=v_x\times t[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} D=(65.25)\times(8.94) \\ =583.33\text{ m} \\ =584\text{ m} \end{gathered}[/tex]
