Answer:
(a) The time of fall is 22.59 s
(b) The final velocity of the present before it hits the ground is 221.38 m/s
Explanation:
Given;
height of fall, h = 2500 m
(A) The time of fall is given as;
[tex]t = \sqrt{\frac{2h}{g} }\\\\t = \sqrt{\frac{2 \ \times \ 2500}{9.8} }\\\\t = \sqrt{\frac{2 \ \times \ 2500}{9.8} }\\\\t = 22.59 \ s[/tex]
(B) The final velocity of the present before it hits the ground is given as;
v = u + gt
where;
u is the initial of fall = 0
v = 0 + (9.8 x 22.59)
v = 221.38 m/s
