A ball is thrown straight up and rises to a maximum height of 22 m. At what height is the speed of the ball equal to half of its initial value

The height of the ball when the speed of the ball is half of its initial value is 16.5 m

Explanation:

Given;

maximum height reached by the ball, h = 22 m

let the initial velocity of the = u

The equation of maximum height reached by the ball is calculated as;

v = u² - 2gh

where;

v is the final velocity at the maximum height = 0

0 = u² - (2 x 9.8)22

0 = u² - 431.2

u² = 431.2

u = √431.2

u = 20.765 m/s

Half of the initial speed = ¹/₂ x 20.765 m/s = 10.383 m/s

The height of the ball at this half value of speed is calculated as;

v² = u² - 2gh

where;

v is the final speed = 10.383 m/s

u is the initial speed = 20.765 m/s

(10.383)² = (20.765)² - (2 x 9.8)h

(2 x 9.8)h = (20.765)² - (10.383)²

19.6h = 323.379

h = 323.379 / 19.6

h = 16.5 m

Therefore, the height of the ball when the speed of the ball is half of its initial value is 16.5 m