The height above ground level in metres of a missile launched vertically, is given by h(t)=-16th^3+100t. At what time is missile 50m in above ground level?
First, fix the typo, and then solve
-16t^2+100t = 50
8t^2 - 50t - 25 = 0
using the quadratic formula.
To find the time at which the missile is 50 meters above ground level, we need to set the height equation equal to 50 meters and solve for t.
The height equation, h(t), is given as:
h(t) = -16t^3 + 100t
Setting h(t) equal to 50 meters:
-16t^3 + 100t = 50
Now, we need to solve this equation for t. Let's break it down step by step.
Step 1: Simplify the equation:
-16t^3 + 100t - 50 = 0
Step 2: Divide both sides by -2 to simplify further:
8t^3 - 50t + 25 = 0
Step 3: At this point, we have a cubic equation. To solve for t, we can either use numerical methods, such as graphing or calculators, or we can attempt to factor the equation.
Factoring can be a bit more challenging with cubic equations, but if we're lucky, there might be a simple root. By trying different small numbers, we can eventually find that t = 1 is a root of the equation.
Step 4: Since t = 1 is a root, we can use polynomial long division or synthetic division to divide the cubic equation by (t - 1):
(8t^3 - 50t + 25) / (t - 1)
Doing the division, we find that the quotient is 8t^2 + 8t - 25.
Step 5: Now, we have a quadratic equation, 8t^2 + 8t - 25 = 0. We can solve this using various methods like factoring, completing the square, or the quadratic formula.
If we use the quadratic formula, which is a general method for solving any quadratic equation of the form ax^2 + bx + c = 0, we have:
t = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 8, b = 8, and c = -25. Plugging these values into the quadratic formula, we get:
t = (-8 ± √(8^2 - 4*8*(-25))) / (2*8)
t = (-8 ± √(64 + 800)) / 16
t = (-8 ± √864) / 16
Simplifying further:
t = (-8 ± √(16 * 54)) / 16
t = (-8 ± 4√54) / 16
t = (-1 ± 0.61) / 2
Considering both the positive and negative solutions, we have two potential values for t:
t1 = (-1 + 0.61) / 2
t1 = -0.39 / 2
t1 = -0.195
t2 = (-1 - 0.61) / 2
t2 = -1.61 / 2
t2 = -0.805
Since time cannot be negative in this context, we discard the negative solution, t = -0.195 seconds.
Therefore, the missile is 50 meters above ground level at approximately t = 0.805 seconds.