AboutJosh Expertise When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.
Experience
Experience: I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.
Supplementary Website: See a selection of past questions in my maths repository under "Question Archive"
Education Credentials: Bachelor degree in Engineering Science."Everyone struggles with something."
Question A model rocket is shot into the air and its path is approximated by h = -5t sqaured + 30t where h is the height of the rocket and t is the elasped time in secounds.
1)when will the rocket hit the ground?
2) what is the maximum height of the rocket?
Answer Hello Ceceli
For 1), you need to find t when the rocket has zero elevation. i.e., you set h=0 in the equation and solve for t.
For 2), the maximum height is found in two steps. Have you learned differentiation yet? If you know how to differentiate any polynomial expression, then you can find the first derivative of h with respect to t. The projectile turns around precisely at the moment when its velocity is zero [motion changes from positive (upward) to negative (downward)]. So, we set dh/dt (viz., the velocity) to 0 and solve for t. Then, we plug in this "t" value into the function h. Here is an illustration.
Step 1: As h=-5t^2+30t, dh/dt=-10t+30. Now we set dh/dt=0. We find -10t+30=0 when t=3.
Step 2: Put t=3 into h=-5t^2+30t. We get 45 (in meters I presume).
