Aball is thrown vertically in the air with a velocity of 100ft/s. use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 150ft

The times are 2.5 sec and 3.75 sec

Step-by-step explanation:

* Lets explain the projectile formula

- The ball thrown vertically in the air with initial velocity 100 ft/sec

- the projectile formula is h = -16 t² + vo t, where h is the height of the

 ball after thrown it in t seconds and vo is the initial velocity

∵ The initial velocity is 100 ft/sec

- substitute vo by 100

∴ The projectile formula is h = -16 t² + 100 t

- To find the time that the ball is at a height of 150 ft substitute

  h by 150 in the projectile formula

∵ h = - 16 t² + 100 t

∵ h = 150 ft

∴ 150 = -16 t² + 100 t

- Subtract 150 from both sides

∴ 0 = -16 t² + 100 t - 150

- Multiply the both sides by -1

∴ 16 t² - 100 t + 150 = 0

* Lets factorize it to find the value of t

∵ 16 t² = 4t × 4t ⇒ first terms in the 2 bracket

∵ 150 = 15 × 10 ⇒ second terms in the two brackets

∵ 4t × 15 = 60 t ⇒ nears

∵ 4t × 10 = 40 t ⇒ extremes

∵ 6t + 40 t = 100 t ⇒ middle term

∴ 16 t² - 100 t + 150 = (4t - 15)(4t - 10)

∵ 16 t² - 100 t + 150 = 0

∴ (4t - 15)(4t - 10) = 0

- Equate each bracket by 0

∴ 4t - 15 = 0 ⇒ add 15 to both sides

∴ 4t = 15 ⇒ divide both sides by 4

∴ t = 3.75

- OR

∴ 4t - 10 = 0 ⇒ add 10 to both sides

∴ 4t = 10 ⇒ divide both sides by 4

∴ t = 2.5

- The ball will be at height 150 ft in 2.5 seconds we the ball goes up

  and again ate 3.75 seconds when the ball goes down after it

  reached its maximum height

* The times are 2.5 sec and 3.75 sec

b. x- 48 =120

step-by-step explanation:

x=6 and x=3

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