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
[tex]h(t) = - 4.9 {t}^{2} + 25t[/tex]
To find the time when the ball hit the ground,we equate the function to zero.
[tex] - 4.9 {t}^{2} + 25t = 0[/tex]
Factor to obtain;
[tex]t( - 4.9t + 25) = 0[/tex]
Apply the zero product property to obtain,
[tex]t = 0 \: or \: \: - 4.9t + 25 = 0[/tex]
[tex]t = 0 \: \: or \: \: t = \frac{ - 25}{ - 4.9} [/tex]
t=0 or t=5.1 to the nearest tenth.
Therefore the ball hits the ground after approximately 5 seconds.
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
