Can you talk me through how to solve problems on projectiles? I always get confused

The strategy is to consider the horizontal and vertical components of the motion separately, and use the equations of motion in each direction. A few things to remember are: the vertical acceleration is always -9.8 ms-2 (due to gravity), the projectile will continue to move upwards until the vertical component of its velocity is zero, the horizontal acceleration is always zero and the horizontal velocity is constant. 

do you remember what the equations of motion are? 

v = u + at, 2as = v2 - u2 and s = ut + ½at2

There are various quantities that you are commonly asked to find.

1) Time to the highest point: use v = u + at vertically, with v = 0

2) Greatest vertical height: use 2as = v2 - u2 vertically, with v = 0, or use s = ut + ½at2 if the time is known

  1. Time taken for it to reach a particular height: use s = ut + ½at2 vertically (if it is returning to the same height at which it started - eg returning to the ground - take s = 0)

  2. Total horizontal distance travelled find the time taken for it to finish its journey (as above), then use s = ut horizontally

Do these rules make sense? Now lets look at some sample questions. 

MK
Answered by Maham K. Physics tutor

2988 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

A satellite is in a stationary orbit above a planet of mass 8.9 x 10^25 kg and period of rotation 1.2 x 10^5 s. Calculate the radius of the satellite's orbit from the centre of the planet.


Explain how bright fringes arise in Young's double slit experiment


A positively charged particle enters a magnetic field oriented perpendicular to its direction of motion. Does the particle: A) Change its velocity, B) Change its speed, C) Accelerate in the direction of the magnetic field.


In the photoelectric effect, what happens as you increase the frequency of light keeping the same intensity constant?


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