Question

H=−4.9t2+25t

Answer 1

ANSWER

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,

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

$t = 0 \: \: or \: \: t = \frac{ - 25}{ - 4.9}$

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

Therefore the ball hits the ground after approximately 5 seconds.

Answer 2

Answer:

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

Step-by-step 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