Answer:
The cliff drivers must leave the top of the cliff at 2.21 m/s
Step-by-step explanation:
Horizontal Motion
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The range or maximum horizontal distance traveled by the object can be calculated as follows:
![\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}](https://img.qammunity.org/2021/formulas/physics/college/h2zuitzheyxmyce4ej5d2wtyr9navlusii.png)
If the range is known and we are required to find the initial speed of launch, then we can solve the above equation for v as follows:
![\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}](https://img.qammunity.org/2021/formulas/physics/high-school/7ne5n0j8ezkcymekbopg2nzevo9354r0t5.png)
Cliff drivers jump into the sea from a height of h=36 m and must pass over an obstacle d=6 meters away from the cliff.
Let's calculate the speed needed:
![\displaystyle v=6~m\cdot\sqrt{\frac {9.8~m/s^2}{2\cdot 36~m}}](https://img.qammunity.org/2021/formulas/physics/college/u6kxei3edxigi70kvh29jpwmg6il0fo4pn.png)
![\displaystyle v=6~m\cdot\sqrt{\frac {9.8~m/s^2}{72~m}}](https://img.qammunity.org/2021/formulas/physics/college/5fi7ojxxrb6fiq3d59poyhps5urvsuhkuf.png)
![v=2.21~m/s](https://img.qammunity.org/2021/formulas/physics/college/dsf981wsmea847x4em9ddaogzyrsv6pujc.png)
The cliff drivers must leave the top of the cliff at 2.21 m/s