Answer:
The velocity of the ball is 3.52 m/s.
Step-by-step explanation:
A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.
Given that the ball was thrown vertically upward on the top of a skyscraper of height 61.9 m. So that the velocity can be determined by;
u =
![\sqrt{(2H)/(g) }](https://img.qammunity.org/2021/formulas/physics/college/i535sey4lvno5fa1zyzkb8bi1pulgam0ye.png)
Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/
, then;
u =
![\sqrt{(2*61.9)/(10) }](https://img.qammunity.org/2021/formulas/physics/college/z6ji2spmxet7k767jaqjfv7zc0gurxi0lb.png)
=
![\sqrt{(123.8)/(10) }](https://img.qammunity.org/2021/formulas/physics/college/xfek3fpfnrqdwcvkyizw25pp4c9dl2xjtn.png)
u = 3.5185
The velocity of the ball is 3.52 m/s.