Question

H=−4.9t2+25t

The equation above expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground?
Question 4 options:

5


4


4.5


3.5

Answer 1

The ball will hit the ground after 5 seconds ⇒ first answer

Explanation:

* Lets study the information in the problem

- The ball is lunched vertically upward from the ground with an initial

velocity 25 meters per second

- The ball will reach the maximum height when its velocity becomes 0

- The ball will fall down to reach the ground again

- The equation of the height (h), in meters of the ball t seconds after

it is lunched from the ground is h = -4.9t² + 25t

- When the ball hit the ground again the height of it is equal 0

∵ h = 0

∴ 0 = -4.9t² + 25t ⇒ Multiply the two sides by -1 and reverse them

∴ 4.9t² - 25t = 0 ⇒ factorize it by taking t as a common factor

∴ t(4.9t - 25) = 0 ⇒ equate each factor by 0

∵ t = 0 ⇒ the initial time when the ball is lunched

∵ 4.9t - 25 = 0 ⇒ add 25 to both sides

∴ 4.9t = 25 ⇒ divide each side by 4.9

∴ t = 5.102 ≅ 5 seconds

* The ball will hit the ground after 5 seconds

Answer 2

5

EXPLANATION

The equation that expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground is

`h(t) = - 4.9 {t}^(2) + 25t`

To find the time when the ball hit the ground,we equate the function to zero.

`- 4.9 {t}^(2) + 25t = 0`

Factor to obtain;

`t( - 4.9t + 25) = 0`

Apply the zero product property to obtain,

`t = 0 \: or \: \: - 4.9t + 25 = 0`

`t = 0 \: \: or \: \: t = ( - 25)/( - 4.9)`

t=0 or t=5.1 to the nearest tenth.

Therefore the ball hits the ground after approximately 5 seconds.