A toy rocket is launched vertically from 5 feet above ground level with an initial velocity of 112 feet per second. The height h after t seconds is given by the equation: h(t) = -16t^2 + 112t + 5
How long will it take for the rocket to reach maximum height? What is the maximum height? and when will it return to the ground.
3 answers
The vertex is 201.2 which would be the maximum height i think
You will have to find the vertex, there are several ways to do this.
1. Use Calculus: find the derivative , set it equal to zero.
that will give you the t of the vertex, sub it back in original to find the height h(t)
h' (t) = -32t + 112 = 0
t = 3.5
h(3.5) = -16(3.5)^2 + 112(3.5) + 5 = 201
so it will take 3.5 seconds to reach a max height of 201 ft
(the vertex is (3.5 , 201)
2. Complete the square. This method you should know, and you should get
h(t) = -16(t-3.5)^2 + 201
3. A good method to learn if you don't know Calculus is this:
for y = ax^2 + bx + c , the x of the vertex is -b/(2a)
so for yours, the t of the vertex = -112/(2(-16)) = 3.5
sub that back into the original and you got your vertex.
1. Use Calculus: find the derivative , set it equal to zero.
that will give you the t of the vertex, sub it back in original to find the height h(t)
h' (t) = -32t + 112 = 0
t = 3.5
h(3.5) = -16(3.5)^2 + 112(3.5) + 5 = 201
so it will take 3.5 seconds to reach a max height of 201 ft
(the vertex is (3.5 , 201)
2. Complete the square. This method you should know, and you should get
h(t) = -16(t-3.5)^2 + 201
3. A good method to learn if you don't know Calculus is this:
for y = ax^2 + bx + c , the x of the vertex is -b/(2a)
so for yours, the t of the vertex = -112/(2(-16)) = 3.5
sub that back into the original and you got your vertex.
well, let me cheat to check
v = dh/dt = -32 t + 112 = 0 at top
so at top t = 112/32 = 3.5 seconds to top (vertex)
so h at top = -16t^2 + 112t + 5
= 5 + 112(3.5) - 16 (3.5)^2
= 201 at the top, ageed
well I told you 3.5 seconds to the top
now for the time when h = 0
0 = -16 t^2 + 112 t + 5
solve quadratic https://www.mathsisfun.com/quadratic-equation-solver.html
7.04 seconds (it only takes a fraction of a second to do that last 5 feet)
v = dh/dt = -32 t + 112 = 0 at top
so at top t = 112/32 = 3.5 seconds to top (vertex)
so h at top = -16t^2 + 112t + 5
= 5 + 112(3.5) - 16 (3.5)^2
= 201 at the top, ageed
well I told you 3.5 seconds to the top
now for the time when h = 0
0 = -16 t^2 + 112 t + 5
solve quadratic https://www.mathsisfun.com/quadratic-equation-solver.html
7.04 seconds (it only takes a fraction of a second to do that last 5 feet)
7 answers