I throw a ball straight up with an initial velocity of 2m/s. How high is it after a fifth of a second?

assign each variable to a value in s = ut + 1/2at2u = 2m/s t=0.2sa=-9.81m/s2so s = (2 x 0.2) + 1/2(-9.81)(0.2)2so s =

HA
Answered by Henry A. Physics tutor

2697 Views

See similar Physics GCSE tutors

Related Physics GCSE answers

All answers ▸

How can an object be at rest without friction?


Why do astronauts feel weightless while in orbit?


Briefly outline how a skydiver reaches terminal velocity.


What is constant acceleration?


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