A rock is thrown straight up with an initial speed of 15.5 m/s . It is caught at the same distance above the ground . How high does the ball rise? How long does the ball remain in the air?

Explanation:

Given;

initial velocity of the rock, u = 15.5 m/s

at maximum height the final velocity of the rock, v = 0

The height risen by the rock is calculated as follows;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (15.5²) / (2 x 9.8)

h = 12.26 m

The time spent in air by the rock is known as time of flight, which is calculated as follows;

v = u - gt

0 = u - gt

gt = u

t = u/g

where;

t is the time to reach maximum height

t = 15.5 / 9.8

t = 1.58 s

Time of flight (time spent in air) is calculated as;

T = 2t

T = 2 x 1.58 s

T = 3.16 s.