Find the time when the object hits the ground. First factor out the GCF of -4, and then factor the equation to find two values of t. The object hits the ground ___ seconds after being thrown. The solution of t = ___ seconds must be rejected. b) The highest point in the objects trajectory occurs halfway between the two times you found in part A. The object is at its highest point ____ seconds after being thrown. c) What is the peak height, the objects height at the time from part b?
An object that is thrown from a cliff follows THIS parabolic trajectory.
The acceleration of gravity determines the quadratic coefficent. When measuring in feet and seconds, we have -16t^2.....
The initial vertical speed determines the linear coefficent. In this case, the object is thrown upwards at 128 feet per second, so we have 128t.
The constant term is determined by the initial height when the object is ghrown. In this case, the object is thrown from a cliff that is 68 feet tall... Hence,
H(t) = -16t^2 + 128t + 68
If we want to find the time when the object crashed into the ground, we wound find the value that makes H(t) = 0
Ex: 0 = -16t^2 + 128t + 68
Questions:
a) Find the time when the object hits the ground. First factor out the GCF of -4, and then factor the equation to find two values of t.
The object hits the ground ___ seconds after being thrown.
The solution of t = ___ seconds must be rejected.
b) The highest point in the objects trajectory occurs halfway between the two times you found in part A.
The object is at its highest point ____ seconds after being thrown.
c) What is the peak height, the objects height at the time from part b?
The object peaks at a height of ___ feet above the ground
d) Imagine a different problem where an object was thrown a speed of 39 feet per second from a cliff that was 53 feet tall.
Write the function H(t), that would describe the objects path as a function of time, t.
Step by step
Solved in 2 steps with 1 images
