A ball is projected vertically upwards from the ground with speed 21 ms^–1. The ball moves freely under gravity once projected. What is the greatest height reached by the ball?

Set out information given in question, and taking the upward direction to be positive: s (displacement) = ?, u (initial speed) = 21ms-1, v (final speed at maximum height) = 0ms-1, a (acceleration when falling freely under gravity) = - 9.8ms-2, t = ?. Using v2 = u2 + 2as: 0 = 212 + 2(-9.8)s, s = 441 / 19.6 = 22.5m. So maximum height reached = 22.5m

SS
Answered by Shruti S. Maths tutor

9085 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve the following equation: 4(sinx)^2+8cosx-7=0 in the interval 0=<x=<360 degrees.


Express 3cos(x)+4sin(x) in the form Rsin(x+y) where you should explicitly determine R and y.


Find the perpendicular bisector passing through the stationary point of the curve y=x^2+2x-7.


For a curve of gradient dy/dx = (2/(x^2))-x/4, determine a) d^2y/dx^2 b) the stationary point where y=5/2 c) whether this is a maximum or minmum point and d) the equation of the curve


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:

© 2026 by IXL Learning