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The maximum height of the projectile will be 226 feet that is, as per question, the option c.
A quadratic equation , y = ax²+bx+c (equation 1)
given, axis of symmetry is
x =
As per the question:
The path of the projectile is modeled using the equation :
h(t) = -16t²+48t+190 (equation 2)
h(t) = height after t time.
comparing equation 1 and equation 2, we get
a = -16 and b = 48
further, t =
=
= 1.5sec
Substituting the found value in equation 2, we get
h(1.5) = -16(1.5)²+48(1.5)+190
h(1.5) = -36+72+190
h(1.5) = 226ft.
Thus, the maximum height of the projectile at 1.5 sec is, 226 feet.
Note that the full question is:
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
To learn more about velocity: brainly.com/question/25749514
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