PedGames Project Curriculum Document
Platform Game Math Worksheet 1
Tags: Player Movement, Speed, Gravity, Equations of Motions
Math Standards: California Standards - Algebra and Functions (AF), Mathematical Reasoning
(MR), Number Sense(NS),
Common Core Standards - High School Algebra – Creating Equations, Reasoning with
Equations & Inequalities
Definitions:
Classic Equations of Motion are given as below:
1) v = u + at
2) s = ut +1/2at2
3) v2 = u2 + 2as
where the symbols are as follows:
v = final velocity
u = initial velocity
s = displacement (distance between initial and final position)
a = acceleration
t = time taken to move from initial to final state.
In Game Maker,
Velocity (u & v) means how many pixels an object moves per step (pixels / step)
Displacement(s) is the no.of pixels.
Acceleration(a) = pixels / step2
Time (t) = no. of steps
Game Maker is set to have 30 steps per second by default.
Hence to calculate time in sec, divide time steps / 30 steps per second = time (sec).
PedGames Project Curriculum Document
Problems:
Use platformtest1.gm81 to check the results and verify the problems.
Example:
Eg: 1) Consider the above world where gravity = 0.5 pixels / step2 and when UP arrow key is
pressed, the ball is thrown upwards with an initial velocity of 12 pixels / step.
Calculate the time required in seconds for the ball to achieve the maximum height before falling
down on the ground. - AF 4.1, MR 1.1, A.CED.1, A-REI.3
Soln:
Given:
Initial velocity u = 12 pixels / step
Final velocity v = 0 (At maximum height, the ball stops, hence v = 0)
Acceleration (gravity) a = -0.5 pixels / step2 (gravity acts downwards whereas ball is thrown
upwards hence negative acceleration)
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Using first equation (v = u + at),
We have,
0 = 12 + (-0.5) x t.
0.5t = 12
t= 24 steps.
Time (sec) = 24 steps / 30 steps per second = 0.8sec
Ans 1: Time required = 0.8 sec for the ball to achieve max height before falling down.
Eg: 2) What will be the total time required for the ball to touch the ground when thrown up with
initial velocity = 12 pixels / step and gravity = 0.5 pixels / step2 – NS 1.2
Soln:
Total time required for the ball to touch ground = 2 * time required to achieve max height
From the above solution
Total time (t) = 2 * 0.8 = 1.6 secs.
Ans 2: Total time required for the ball to touch ground = 1.6 secs.
Questions:
Q1) Consider the gravity = 0.5 pixels / step2 the ball is thrown upwards with an initial velocity of
15 pixels / step. Calculate the time required in seconds for the ball to achieve the maximum
height before falling down on the ground - AF 4.1, MR 1.1, A-CED.1, A-REI.3
PedGames Project Curriculum Document
Q2) What will be the total time required for the ball to touch the ground when thrown up with
initial velocity = 15 pixels / step and gravity = 0.5 pixels / step2 - NS 1.2
Eg: 3) Given gravity = 0.5 pixels / step2, time to attain maximum height for the ball = 1 sec,
calculate the initial velocity with which the ball was thrown - AF 4.2, MR 1.1, A-CED.1, A-REI.3
Solution:
Given: gravity a = -0.5 pixels / step2,
v = 0 pixels / step. (At maximum height, the ball stops, hence final velocity = 0)
t = 1sec.
To find: u =?
First calculate time in terms of steps:
Time (steps) = 1 sec x 30 steps / sec = 30 steps.
Using first equation (v = u + at),
0 = u + -0.5 x 30
u = 15 pixels / step.
Ans 3: The initial velocity u = 15 pixels / step with which the ball was thrown
PedGames Project Curriculum Document
Eg: 4) Calculate the distance that the ball travels when it reaches the maximum height when it
is thrown with a velocity of 6 pixels / step and gravity of the world is set to 0.2 pixels / step2 –
Algebra I 23.0, MR 1.1, A-CED.1, A-REI.3
Soln:
Given:
Initial velocity u = 6 pixels / step
Final velocity v = 0 pixels / step (At max height , the ball stops hence v = 0).
Acceleration a = -0.2 pixels / step2
To find: s = ?
Using the third equation of motion (v2 = u2 + 2as), we have
0 = 62 + 2 x -0.2 x s
-36 = -0.4 x s
s = 36 / 0.4 = 360 / 4 = 90 pixels.
Ans 4: Distance covered by the ball = 90 pixels
Questions:
Q3) Calculate the velocity at which the ball is thrown up when it attains a maximum height at 2
sec and gravity = 0.4 pixels / step2.
Q4) Calculate the total distance covered by the ball when it is thrown up and back to the ground
with a velocity of 10 pixels / step and gravity set to 0.2 pixels / step2.
PedGames Project Curriculum Document
Eg: 5) As shown in the above figure, there is a brick wall which is made up of 8 bricks, each 32
pixels wide and 32 pixels high. The bricks are placed on top of one another to make the wall.
So at what velocity should the ball be thrown up so that it can jump the wall given acceleration
due to gravity is 0.5 pixels / step2. AF 4.2, MR 1.1, 1.3, Algebra I 23.0, A-CED.1, A-REI.3
Given:
Gravity a =0.5 pixels / step2.
Final Velocity at max height v = 0 pixels / step.
Distance s is given in terms of height of the wall.
For the ball to jump across the wall, the ball should travel the no. of pixels as the height of the
wall.
Wall Height = 8 x 32 = 256 pixels.
Hence s = 256 pixels.
Using third equation of motion (v2 = u2 + 2as).
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0 = u2 + 2 x -0.5 x 256
u2 = 256
u = 16 pixels / step.
Ans 5: The ball should be thrown at least 16 pixels / step if it needs to jump the wall of
the given height.
Question:
Q5) Given that the maximum speed with which the ball can be thrown is 12 pixels / step and
gravity set to 0.5 pixels / step2, what should be the minimum height wall that is needed so that
the ball cannot jump. Also calculate the minimum no. of bricks required to make such wall given
that dimensions of each brick is 36 pixels wide and 36 pixels high.
AF 4.2, MR 1.1, 1.3, Algebra I 23.0, A-CED.1, A-REI.3
PedGames Project Curriculum Document
Eg: 6) A ball is falling from a height of 150 pixels and it touches the ground with a velocity 30
pixels/sec. Calculate the acceleration due to gravity. MR 1.1, Algebra I 23.0, A-CED.1, A-REI.3
Soln:
Given:
Initial Velocity u = 0. (The ball is starting to fall from a height, hence its initial speed = 0)
Final Velocity v = 30 pixels /sec
Displacement s = 150 pixels.
To find: a = ?
Using third equation of motion (v2 = u2 + 2as),
302 = 0 + 2 x a x 150
900 = 300a
a = 3 pixels / step2
Ans 6: Acceleration due to gravity = 3 pixels/step2.
Eg: 7) A ball falling from a height reaches the ground in 2 secs. Given that gravity = 0.5
pixels/step2, calculate the height from which the ball falls. AF 4.2, MR 1.1, 1.3, A-CED.1, AREI.3
Soln:
Given:
Initial Velocity u =0. (The ball is starting to fall from a height, hence its initial speed = 0)
Time t = 2 sec
a = 0.5pixels/step2
Time in no.of steps = 2 sec x 30 steps / sec = 60 steps.
Using second equation of motion( s = ut + 1/2at2)
s = 0 x 60 + ½ x 0.5 x 602
s = 0 + ¼ x 3600
s = 900 pixels
PedGames Project Curriculum Document
Ans 7: Height = 900 pixels from where the ball starts to fall.
Questions
Q6) A ball rolls from a ledge that is at a height of 200 pixels and it reaches the ground with a
velocity of 20 pixels / step. Calculate the acceleration due to gravity attained by the ball when it
reaches the ground? MR 1.1, 1.3, Algebra I 23.0, A-CED.1, A-REI.3
Q7) A ball takes 4 sec to reach the ground with a velocity of 24 pixels / step. Calculate the
acceleration due to gravity? AF 4.2, MR 1.1, 1.3, A-CED.1, A-REI.3