A projectile is launched from a platform 10 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 + 10 feet.
To find the maximum height reached by the projectile, we need to determine the vertex of the quadratic equation. The vertex form of a quadratic equation is given by:
y = a(x - h)^2 + k
In this case, the equation is h = -16t^2 + 72t + 10. Comparing this with the vertex form, we can find the maximum height.
1. First, we need to determine the value of t at the vertex. The formula for finding the x-coordinate of the vertex is given by x = -b / (2a), where a = -16 and b = 72 in this case.
Using the formula, we can calculate t = -72 / (2 * -16) = 2.25 seconds.
2. Now, substitute t = 2.25 into the equation h = -16t^2 + 72t + 10 to find the maximum height.
h = -16(2.25)^2 + 72(2.25) + 10
= -16(5.0625) + 162 + 10
= -81 + 162 + 10
= 91 feet (rounded to the nearest whole number)
Therefore, the maximum height reached by the projectile is 91 feet.
To determine when the projectile will reach its maximum height, we need to find the vertex of the quadratic equation for the height.
The equation for the height of the projectile is given by h = –16t^2 + 72t + 10 feet.
The vertex of a quadratic equation in the form of h = ax^2 + bx + c can be found using the formula:
t = -b / (2a)
In this case, a = -16 and b = 72.
Substituting these values into the equation, we find:
t = -72 / (2 * -16)
t = -72 / -32
t = 2.25
Therefore, the projectile will reach its maximum height at 2.25 seconds after launch.