A ball is thrown in the air. The height of the ball at time t is given by: h=5+4t-2t^2. What is its maximum height? At what time does the ball reach this height?

First, we find the derivative of h: dh/dt= 4-4t. To find the point(s) of interest, we solve dh/dt=0. This gives the answer t=1. In order to determine whether t=1 is a minimum point or maximum point we find the second derivative of h: d2h/dt2=-4. As the second derivative of h is less than 0, this shows that there is a maximum point at t=1. Therefore, the ball reaches its maximum height when t=1. To determine the maximum height, we substitute t=1 into the equation for h. Here, we find the maximum height achieved by the ball is h=7.

DS

Related Maths A Level answers

All answers ▸

find the integral of f'(x)=2x+5


The equation x^3 - 3*x + 1 = 0 has three real roots; Show that one of the roots lies between −2 and −1


A function is defined parametrically as x = 4 sin(3t), y = 2 cos(3t). Find and simplify d^2 y/dx^2 in terms of t and y.


Given that y= 5x^2 + 2x , find dy/dx