Answer:
The stone will be
above the ground in
![2.3s](https://img.qammunity.org/2023/formulas/physics/high-school/275p4tft67st8zqr1veah5devzx5sei4vz.png)
Step-by-step explanation:
As the stone is thrown vertically upward from ground with speed i.e.,
the initial velocity,
,
the final velocity,
,
acceleration due to gravity,
(upward motion)
![v=u+at\\0=25+(-10)t\\0=25-10t\\t=2.5s](https://img.qammunity.org/2023/formulas/physics/high-school/rv4uao7hxcqixo2pvv0et3j8e48jttwgdy.png)
To calculate the height at which the stone reached above the ground we can use the below formula:
![s=ut+(1)/(2)at^2](https://img.qammunity.org/2023/formulas/physics/college/obvwlaq2brhvduordk5kuedw6274qqa7x1.png)
![s=62.5-5*6.25=62.5-31.25=31.25m](https://img.qammunity.org/2023/formulas/physics/high-school/uawr1r00nu790fqctlxbf9f3w6jitnrcsg.png)
Now, the time taken by stone to reach
above the ground can be calculated using
![u=0m/s\\s=31.25-5=26.25m\\a=10m/s^2\\t=?](https://img.qammunity.org/2023/formulas/physics/high-school/81k7cma5lg0ychgbi34rwru83zfhqrxauh.png)
![s=ut+(1)/(2) at^2\\](https://img.qammunity.org/2023/formulas/physics/high-school/sopqa0k2i2x75o20jivifd1o0gkwkk5fkd.png)
![26.25=0 * t+(1)/(2) * 10 * t^2](https://img.qammunity.org/2023/formulas/physics/high-school/umeuzfmnoj870v6zg1x0rx8phzy8qxtwjl.png)
![26.25=0+5t^2](https://img.qammunity.org/2023/formulas/physics/high-school/tp4mp0dwz1t31s02jyahcftl0mtodepdeq.png)
![5t^2=26.25](https://img.qammunity.org/2023/formulas/physics/high-school/4n00eiv6wl914q7tm53rn4nt4x1fkzj70k.png)
![t^2=(26.25)/(5)=5.25](https://img.qammunity.org/2023/formulas/physics/high-school/642df07qp95hdbqsasvpjmpkbzmrddulr1.png)
![t=√(5.25) =2.3s](https://img.qammunity.org/2023/formulas/physics/high-school/adbqvdafo4b0me7l4fgfe1vlncmw5zobv8.png)
Thus, at
the stone will reach
above the ground.