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

4451 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

When using the trapezium rule to approximate area underneath a curve between 2 limits, what is the effect of increasing the number of strips used?


Integrate ((5x^3) + ((2x)^-1) + (e^2x))dx.


find the derivative of f(x) = x^3 + 2x^2 - 5x - 6. Find all stationary points of the function.


If n is an integer prove (n+3)^(2)-n^(2) is never even.


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