Question

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

Answer 1

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

Step-by-step 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