Projectile Motion
Brook Edgar
Teacher
Contents
Explainer Video
SUVAT's
SUVAT equations are a set of equations used whenever the acceleration is constant.
The equations can be derived from graphs.
In a graph, the acceleration is the gradient of the line.
Formula:
In a graph, the displacement is the area under the graph.
Formula:
The other equations of motion can be derived using these first two. They are all given on the formula sheet.
Worked Example
A skateboarder travelling in a straight line up a ramp experiences a constant deceleration . His velocity at is .
How long does it take for the skateboarder to reach his maximum height up the ramp?
How high up the ramp can he go?
Answer:
Always begin by setting up your SUVAT. Then think which equation of motion is the easiest to use to solve for the unknown.
Worked Example
A particle is moving in a straight line with constant acceleration, . At time , the speed of the particle is .
Calculate the speed of the particle at .
Calculate its displacement at .
Answer:
Always begin by setting up your SUVAT. Then think which equation of motion is the easiest to use to solve for the unknown.
Worked Example
A driver wants to accelerate his car uniformly from to . He has of track to do so. Calculate the constant acceleration needed from the car.
Answer:
Always begin by setting up your SUVAT. Then think which equation of motion is the easiest to use to solve for the unknown.
Projectile Motion
SUVATs are used in projectile motion, as we can assume a constant acceleration in the vertical direction, as air resistance is ignored. This constant acceleration of downwards is due to the force of gravity, causing the vertical velocity to increase at a constant rate. The horizontal speed remains constant, as we have ignored air resistance.
Here we can see that the horizontal velocity of the projectile remains constant while the vertical velocity increases at a constant rate.
Remember, velocity is a vector, so we regard the downwards direction as negative.
We can use SUVATs for objects in free fall as the only force acting on the object when in free fall is that due to gravity. As acceleration is a vector, the value we use for acceleration is .
Another key fact to remember is that when an object is projected upwards, its vertical velocity is zero at maximum height, as at this point the object is changing direction. However, its horizontal velocity remains constant as we are ignoring air resistance.
For example, to calculate the maximum height reached by a ball projected vertically upwards from the ground at , we set up our SUVATS, knowing that at maximum height, the speed is zero and that the acceleration due to gravity is constant.
Worked Example
A ball is projected vertically upwards from a point, above the ground. The initial velocity is upwards. How long does it take for the ball to hit the ground?
Answer:
Always begin by setting up your SUVAT. In this question we know that the final displacement of the ball will be , as the ball ends up nine metres below its starting point.
To calculate time, we could use the quadratic formula to solve the equation of motion;
However, this can be time-consuming, so we can solve for time in two separate parts.
Teacher Tip: Square roots have two solutions, a negative and a positive, if we look at our image the final velocity is downwards , so we want the negative solution.
Projectiles at an Angle
To solve problems like the one below, where an athlete throws a javelin at at angle of to the horizontal, we need to resolve the velocity vector into its vertical and horizontal components, as they are independent of each other in order to calculate the time of flight and range of the throw.
To simplify the throw we can draw the vector diagram below,
We can use SOHCAHTOA to determine the horizontal and vertical components, or you can remember the rule that if the vector passes through the angle, it is always cosine, and if you do not pass through the angle when resolving, then it is sine.
Vertical component =
Horizontal component =
We can now set up our SUVAT to determine the time of flight.
-> As the object returns to the ground at the end of motion, the vertical displacement is zero metres.
To find the range of the throw, we need to use the horizontal component of the velocity. As air resistance is ignored, the horizontal speed is constant, we can use the equation, speed = distance time.
Remember: SOHCAHTOA is used to remember the trigonometric ratios.
Worked Example
A man throws a stone horizontally from the top of a cliff at .
Calculate the time it takes for the stone to reach the bottom of the cliff.
Calculate how far horizontally the stone reaches from the bottom of the cliff.
Determine the velocity and direction it hits the ground at.
Answer:
Always begin by setting up your SUVAT. In this question, we know that the initial vertical velocity is zero as the stone is thrown horizontally, and that's its final displacement is fifty metres below where it started.
We first had to calculate the final velocity of the object in the vertical direction, as the velocity in the vertical direction increases over time. We can then use trig to find the final velocity as the horizontal component of the velocity is constant, as air resistance is ignored.
Pythagoras' theorem, .
below the horizontal.
Practice Questions
A cannonball is fired horizontally from the top of a castle towards an enemy ship, as shown below,
Show that the time taken for the cannonball to reach the ship is .
Calculate the magnitude and direction of the cannonball just before it hits the ship.
-> Check out Brook's video explanation for more help.
Answer:
below the horizontal.
A ball is kicked horizontally at from the edge of a high hill.
Calculate:
Time of flight
Range
-> Check out Brook's video explanation for more help.
Answers:
Time
Range
A projectile is launched at a speed of at an angle to the horizontal.
Find the time for the projectile to return to the ground.
-> Check out Brook's video explanation for more help.
Answer: