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

4300 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate x^2e^x with respect to x between the limits of x=5 and x=0.


f(x) = x^3 - 13x^2 + 55x - 75 , find the gradient of the tangent at x=3


(a) By using a suitable trigonometrical identity, solve the equation tan(2x-π/6)^2 =11-sec(2x-π/6)giving all values of x in radians to two decimal places in the interval 0<=x <=π .


Find the equation for the tangent to the curve y^3 + x^3 + 3x^2 + 2y + 8 = 0 at the point (2,1)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences