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A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = –16t2 + 48t + 190. What is the maximum height of the projectile? 82 feet 190 feet 226 feet 250 feet

Respuesta :

Answer:

226 feet

Step-by-step explanation:

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The maximum height of the projectile is 226 feet

How to calculate the maximum height of an object?

Given the equation of the height modelled as h(t) = –16t^2 + 48t + 190

The ball reaches the maximum height at the point where v = dh/dt = 0

dh/dt = -32t + 48
32t = 48
t = 1.5s

Get the maximum height;

h(1.5) = –16(1.5)^2 + 48(1.5) + 190

h(1.5) = -36 + 72 + 190
h(1.5) = 226 feet

Hence  the maximum height of the projectile is 226 feet

Learn more on maximum height here: https://brainly.com/question/12446886

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