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
Answered by Debbie S. Maths tutor

4394 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Using the substitution u = 2 + √(2x + 1), or other suitable substitutions, find the exact value of 4 0 1 ∫ 2 (2 1) +√ +x dx giving your answer in the form A + 2ln B, where A is an integer and B is a positive constant


Find the equation of the normal of the curve xy-x^2+xlog(y)=4 at the point (2,1) in the form ax+by+c=0


Find dy/dx in terms of t for the curve given by the parametric equations x = tan(t) , y = sec(t) for -pi/2<t<pi/2.


Simplify: (3x+8)/5 > 2x + 1


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences