A person standing near the edge of a cliff 120 feet above a lake throws a rock upward with an initial speed of 40 feet per second. The height of the rock above the lake at the bottom of the cliff is a function of time and is described by the function H(t) = –16t2 + 40t + 120.
a) How many seconds will it take until the rock reaches its maximum height?
b) What is that height?
H(t) = -16t^2 + 40t + 120
Finding critical point...
H'(t) = -32t + 40 ...... Differentiation
0 = -32t + 40
32t = 40
t = 1.25 seconds
Thats you answer for (a). Use it to find the answer for (b).
Thats you...
I meant "that's your..."
a) To find the time it takes for the rock to reach its maximum height, we need to determine when the velocity of the rock changes from going upward to going downward. The velocity of the rock can be found by taking the derivative of the function for the height:
V(t) = dH/dt = -32t + 40
We set V(t) equal to zero and solve for t:
-32t + 40 = 0
-32t = -40
t = 40 / 32
t = 1.25 seconds
Therefore, it will take the rock 1.25 seconds to reach its maximum height.
b) To find the maximum height of the rock, we substitute the time value we found in part a) into the height function H(t):
H(1.25) = -16(1.25)^2 + 40(1.25) + 120
H(1.25) = -16(1.5625) + 50 + 120
H(1.25) = -25 + 50 + 120
H(1.25) = 145 feet
Therefore, the maximum height of the rock above the lake is 145 feet.
To find the answer to the first question, we need to determine the time it takes for the rock to reach its maximum height.
a) The maximum height of the rock occurs at the vertex of the quadratic equation H(t) = -16t^2 + 40t + 120. The vertex of a quadratic equation in the form of H(t) = at^2 + bt + c is given by the formula t = -b / (2a).
In our case, a = -16 and b = 40. Plugging these values into the formula, we get:
t = -40 / (2 * -16)
t = -40 / -32
t = 1.25
So, it will take 1.25 seconds for the rock to reach its maximum height.
b) To find the maximum height, we substitute the value of t = 1.25 into the equation H(t) = -16t^2 + 40t + 120.
H(1.25) = -16(1.25)^2 + 40(1.25) + 120
H(1.25) = -16(1.5625) + 50 + 120
H(1.25) = -25 + 50 + 120
H(1.25) = 145
Therefore, the maximum height of the rock above the lake is 145 feet.