Therefore there will be a?maximum?height?that the projectile reaches
At the maximum height
Horizontal?velocity,?vx?m s-1, will be the same as the initial horizontal velocity?ux
Vertical?velocity,?vy?m s-1?will be instantaneously zero
If the projectile?launches and lands?at the same height then the?horizontal distance?to the?maximum height?is?half?of the?range of the projectile
Time?t?is the parameter that will be the same for both the horizontal and vertical components
Harder concepts with projectiles
?You will need all your skills from your previous work with?suvat
Speed is the?magnitude?of velocity and can be found by using Pythagoras’ Theorem
A good understanding of the relationship between?distance?and?displacement
Simultaneous equations
Finding the point of intersection of two projectiles
Sometimes the launch and landing points will be at different heights such as if a stone is thrown from the top of a cliff into the sea
Think carefully about the vertical displacement from the start to the end
You might get asked to use the model of a projectile to decide whether an object passes over a certain height
Find the time taken to travel the horizontal distance and use this to find the maximum vertical height of the projectile
Worked Example
(a)? Find the maximum height above the ground that the stone reaches.
(b)? Find the time taken for the stone to hit the ground.
(c)? Find the horizontal distance from the building to the point where the stone lands.
Exam Tip
Always?draw a diagram and add to it as you work through the question. Make it clear which direction you are using as positive.
Questions could ask you to find the speed at an instant, to do this remember to first find both components of the velocity and then find the magnitude of the velocity.
Not all projectiles are projected upwards, for example a cannonball fired from the turret of a castle aimed at an enemy on the ground!
Remember to round answers to 2 significant figures if g=9.8 is used.