Given,
The initial velocity of the projectile, u=128 ft/s
The acceleration due to gravity, g=-32 ft/s²
After reaching the maximum height the velocity of the projectile will be v=0 m/s
The time(t₁) required for the projectile to reach the maximum height is given by,
[tex]v=u+gt_1[/tex]On substituting the known values,
[tex]\begin{gathered} 0=128+(-32)\times t_1 \\ t_1=\frac{-128}{-32} \\ =4\text{ s} \end{gathered}[/tex]The maximum height (h) reached by the projectile is given by,
[tex]v^2-u^2=2gh[/tex]On substituting the known values,
[tex]\begin{gathered} 0-128^2=2\times-32\times h \\ h=\frac{-128^2}{2\times-32} \\ =256\text{ ft} \end{gathered}[/tex]The time (t₂) required for the projectile to come back to the ground from its maximum height is calculated using the formula,
[tex]h=vt_2-16t^2_2[/tex]Here the value of the height will take a negative value because the projectile will be covering that distance when it is falling downwards
On substituting the known values,
[tex]\begin{gathered} -256=0-16t^2_2 \\ \Rightarrow t_2=\sqrt{\frac{256}{16}} \\ =4\text{ s} \end{gathered}[/tex]Thus the total time in which the projectile returns to the ground is
[tex]\begin{gathered} t=t_1+t_2 \\ =4+4=8\text{ s} \end{gathered}[/tex]A) Thus the projectile will return to the ground in 8 s
B) The maximum height reached by the projectile is 256 ft
C)
The height, h=240 ft
The time(t₃) required to reach this height is given by,
[tex]h=ut_3-16t^2_3[/tex]On substituting the known values,
[tex]\begin{gathered} 240=128\times t_3-16t^2_3 \\ \Rightarrow16t^2_3-128t_3+240=0 \end{gathered}[/tex]On solving the above equation,
[tex]t_3=5_{}\text{ s}[/tex]And
[tex]t_3=3\text{ s}[/tex]Thus the projectile will reach the height at 3s when going up and again at 5s when coming down.
