Answer:
<h3>The option C) 15 s is correct</h3><h3>Therefore the maximum height it reaches in <u>15 seconds. </u></h3>
Step-by-step explanation:
Given equation is 
Given that a projectile is thrown upward so that its distance above the ground after t seconds is 
<h3>To find how many seconds does it reach its maximum height:</h3>
"The standard form of a parabola's equation is expressed as :
.
If
, then the parabola opens upwards;
if
the parabola opens downwards."
The maximum height is the vertex of the parabola
which is
.
Comparing the given equation with the standard form of parabola we get
the values of a=-13 , b=390 and c=0
The maximum height in
.
Substitute the values we get



<h3>∴ t=15 s</h3><h3>∴ The option C) 15 s is correct.</h3><h3>∴ the maximum height is the vertex of the parabola, at t reaches in 15 seconds. </h3>
Answer:
from left to right, including the triangle placed in the rectangle section already; quadrilateral, polygon, quadrilateral, polygon, quadrilateral, rectangle
<span>We have a right angled triangle with an opposite of 300.5 ft. (306 - 5.5) and an adjacent of 400 ft. Recalling SOH CAH TOA, tanθ = O/A.
tan(θ) = 300.5/400.
θ = tan^-1(300.5/400).
θ = 36.9°.</span>