A projectile is launched from a platform 20 feet high with an initial velocity of 72 feet per second,
The height h of the projectile at t seconds after launch is given by h = –16t2 + 72t + 20 feet.
(a) How many seconds after launch does the projectile attain maximum height? Show work.
(b) What is the maximum height? Show work.
To find the maximum height of the projectile and the time it takes to reach it, we need to analyze the equation for height h = -16t^2 + 72t + 20.
(a) The time at which the projectile attains its maximum height can be determined by finding the vertex of the parabolic equation. The x-coordinate of the vertex is given by -b/(2a), where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
In our case, a = -16 and b = 72. Substituting these values into the formula, we get:
t = -72 / (2 * -16)
t = -72 / -32
t = 2.25 seconds
Therefore, the projectile attains its maximum height 2.25 seconds after launch.
(b) To find the maximum height, we substitute the value of t = 2.25 back into the equation for h:
h = -16(2.25)^2 + 72(2.25) + 20
h = -16(5.0625) + 162 + 20
h = -81 + 162 + 20
h = 101 feet
So, the maximum height of the projectile is 101 feet.
In summary:
(a) The projectile attains its maximum height 2.25 seconds after launch.
(b) The maximum height of the projectile is 101 feet.