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A projectile is launched from ground level with an initial velocity of v 0 feet per second. Neglecting air​ resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v 0 t. Find the​ time(s) that the projectile will​ (a) reach a height of 288 ft and​ (b) return to the ground when v 0equals144 feet per second. ​(a) Find the​ time(s) that the projectile will reach a height of 288 ft when v 0equals144 feet per second. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

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Answer:

After 3 seconds while going up and 6 seconds while coming down

Step-by-step explanation:

Since projectile is launched from ground level with an initial velocity of v 0 feet per second, initial height =0

Height is given by

[tex]h(t) = -16t^2+144t[/tex] where v0 = initial velocity = 144

a) When h =288 ft.

[tex]288 = -16t^2+144t\\t^2-9t+18 =0\\t=6 or 3[/tex]

At t= 3 or 6 seconds the projectile would be at height 288 ft.

So while going up after 3 seconds it wouldbe at a height of 288 ft and while coming after 6 seconds.

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