So, the final velocity of the ball when it is 2.101 above the ground is 13.99 m/s or can be rounded to 14 m/s.
Hi ! In this question, I will help you. This question will addopt the principle of final velocity in free fall. Free fall is vertical downward movement that occurs when any object dropped without initial velocity. In othee word, the object that falls is only affected by the presence of gravity and its initial high. In general, the final velocity in free fall can be expressed by this equation :
[tex] \boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}} [/tex]
With the following condition :
We know that :
Note :
At this point (2.101 m above the ground), the object can still complete its movement up to exactly 0 m above the ground.
What was asked :
Step by Step
[tex] \sf{v = \sqrt{2 \times g \times \Delta h}} [/tex]
[tex] \sf{v = \sqrt{2 \times g \times (h_1 - h_2)}} [/tex]
[tex] \sf{v = \sqrt{2 \times 9.8 \times (12.09 - 2.101)}} [/tex]
[tex] \sf{v = \sqrt{19.6 \times 9,989}} [/tex]
[tex] \sf{v \approx \sqrt{195.78}} [/tex]
[tex] \boxed{\sf{v = 13.99 \: m/s \approx 14 \: m/s}} [/tex]
So, the final velocity of the ball when it is 2.101 above the ground is 13.99 m/s or can be rounded to 14 m/s.
