A small stone is projected verically upwards from a point O with a speed of 19.6ms^-1. Modeeling the stone as a particle moving freely under gravity find the time for which the stone is more than 14.6m above O

S = 14.7, U = 19.6, V =,  A = -g, T = t

using s = ut + 1/2 at^2
14.7 = 19.6t + 1/2 -g t^2
1/2 g t^2 - 19.6t + 14.7 = 0

t = (19.6 +- sqrroot(-19.6- 4 * 0.5 * 9.8 * 14.7)) / 2 * 0.5 * 9.8

t = 1 and t = 3

Therefore the total time above 14.7 was 2 seconds

HB
Answered by Hamish B. Maths tutor

4593 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I integrate terms with sin^2(x) and cos^2(x) in them? For example integrate (1+sin(x))^2 with respect to x


What is a derivative and how do we calculate it from first principles?


Find the integral of ln x


Integrate sin7xcos3x


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning