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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?
A. 82 feet
B. 190 feet
C. 226 feet
D. 250 feet

2 Answers

7 votes

The answer is C on edge

User Alfredo Morales
by
7.9k points
5 votes

Answer:

Option c is correct

226 feet is the maximum height of the projectile.

Explanation:

A quadratic equation
y=ax^2+bx+c, ....[1]

then the axis of symmetry is given by:


x = -(b)/(2a)

As per the statement:

The path of the projectile is modeled using the equation :


h(t) = -16t^2+48t+190 ....[2]

where, h(t) is the height after t time.

On comparing with [1] we have;

a = -16 and b = 48

then;


t = -(48)/(2(-16)) = (48)/(32) = 1.5 sec

Substitute this in [2] we have;


h(1.5) = -16(1.5)^2+48(1.5)+190


h(1.5) = -36+72+190

Simplify:


h(1.5) = 226 ft

Therefore, the maximum height of the projectile at 1.5 sec is, 226 feet.

User Ansal Ali
by
8.5k points
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