Answer:
[tex]t=2.98s[/tex]
Explanation:
First, we have to calculate the maximum height reached by the object. We use the equations of uniformly accelerated motion:
[tex]y=y_0+v_0t-\frac{gt_1^2}{2}[/tex]
In order to calculate this, we have to know the time taken by the object to reach the maximum height:
[tex]v_f=v_0-gt_1\\0=32\frac{ft}{s}-(32.15\frac{ft}{s^2})t_1\\t_1=\frac{-32\frac{ft}{s}}{-32.15\frac{ft}{s^2}}\\t_1=0.99s[/tex]
Now, we can calculate y:
[tex]y=48ft+(32\frac{ft}{s})0.99s-\frac{(32.15\frac{ft}{s^2})(0.99s)^2}{2}\\y=63.93ft[/tex]
Now, we calculate the time taken by the object in free fall:
[tex]y=y_0+v_0t+\frac{gt_2^2}{2}\\y=0ft+0\frac{ft}{s}t+\frac{gt_2^2}{2}\\t_2=\sqrt\frac{2y}{g}\\t_2=\sqrt\frac{2(63.93ft)}{32.15\frac{ft}{s^2}}\\t_2=1.99s[/tex]
Finally, adding [tex]t_1[/tex] and [tex]t_2[/tex], we get the total time until the object impacts the ground:
[tex]t=t_1+t_2\\t=0.99s+1.99s\\t=2.98s[/tex]
