Question
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.
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.
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)
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