Answer:
![t=2.98s](https://img.qammunity.org/2021/formulas/physics/high-school/xvsfu7lcaxkh7lyaa4t2465201d8adcm98.png)
Step-by-step explanation:
First, we have to calculate the maximum height reached by the object. We use the equations of uniformly accelerated motion:
![y=y_0+v_0t-(gt_1^2)/(2)](https://img.qammunity.org/2021/formulas/physics/high-school/g16rls21vkzqm466appm1gt0sr85ct373r.png)
In order to calculate this, we have to know the time taken by the object to reach the maximum height:
![v_f=v_0-gt_1\\0=32(ft)/(s)-(32.15(ft)/(s^2))t_1\\t_1=(-32(ft)/(s))/(-32.15(ft)/(s^2))\\t_1=0.99s](https://img.qammunity.org/2021/formulas/physics/high-school/atfc3z5xzhlq5zpkq1yy6d1zugy0xva5om.png)
Now, we can calculate y:
![y=48ft+(32(ft)/(s))0.99s-((32.15(ft)/(s^2))(0.99s)^2)/(2)\\y=63.93ft](https://img.qammunity.org/2021/formulas/physics/high-school/nlmwn6clf35xwnx4h5qcaufkhon0mby3ao.png)
Now, we calculate the time taken by the object in free fall:
![y=y_0+v_0t+(gt_2^2)/(2)\\y=0ft+0(ft)/(s)t+(gt_2^2)/(2)\\t_2=\sqrt(2y)/(g)\\t_2=\sqrt(2(63.93ft))/(32.15(ft)/(s^2))\\t_2=1.99s](https://img.qammunity.org/2021/formulas/physics/high-school/s745w7s5e8gw0ho8tk5muucsfeagvtybnp.png)
Finally, adding
and
, we get the total time until the object impacts the ground:
![t=t_1+t_2\\t=0.99s+1.99s\\t=2.98s](https://img.qammunity.org/2021/formulas/physics/high-school/dfiwy5e4h8nsolzuxb9y9k4sbkbo2q07am.png)