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The height h, in feet, of a projectile t seconds after launch is modeled by the equation h = 32t – 16t2. how long after launch does the projectile return to the ground? 2 seconds4 seconds14 seconds16 seconds

Respuesta :

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dy/dx h = 32t - 16t^2
v = 32 - 32t
at Max hight speed is 0m/s
therefore

0 = 32 - 32t
t = 32/32 = 1 s

so it goes up to Max hieght in 1 s
and it takes another 1s to come back
thus time after launch to return to ground is 2s!

Answer:

The correct option is A. 2 seconds

Step-by-step explanation:

The height h, in feet, of a projectile t seconds after launch is modeled by the equation h = 32·t – 16·t²

Now, we need too find the time after which the fired projectile touches the ground.

So, The fired projectile when touches the ground, The height of the projectile will become 0

⇒ Find the required time t when h = 0

⇒ 0 = 32·t - 16·t²

⇒ 0 = t × (32 - 16·t)

⇒ t = 0 or 32 - 16t = 0

⇒ 16t = 32

⇒ t = 2

So, t = 0 or t = 2

t cannot be 0 because initially the projectile is in the air as it has been already fired

⇒ t = 2 seconds

Therefore, The correct option is A. 2 seconds

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