Explanation:
It is given that,
Height of the building, h = 44 m
Initial horizontal velocity, [tex]u_x=ucos\theta=8.6\ m/s[/tex]
Initial vertical velocity, [tex]u_y=usin\theta=10.5\ m/s[/tex]
It can be assumed to find the maximum height of the projectile and time taken to reach the maximum height.
The formula for the maximum height is given by :
[tex]H=\dfrac{(u\ sin\theta)^2}{2g}[/tex]
g is the acceleration due to gravity
[tex]H=\dfrac{(10.5)^2}{2\times 9.8}=5.625\ m[/tex]
The total maximum height, h' = h + H
[tex]h'=44+5.625=49.625\ m[/tex]
The formula for the time of flight is given by :
[tex]t_{max}=\dfrac{u\ sin\theta}{g}[/tex]
[tex]t_{max}=\dfrac{10.5}{9.8}=1.07\ s[/tex]
Hence, this is the required solution.
