Projectile Motion.
Ever think how you could throw a ball and know where it is going to land? With the help of the Projectile Motion Unit, We are capable of getting calculating such tasks with equations with a combination of what we've learned so far with acceleration and velocity.
I will be demonstrating
Projectile 1 - I can represent the initial and final velocities of a projectile as vectors.
Projectile 2 - I can calculate the x and y components of a projectile's initial velocity.
Projectile 3 - I can calculate the necessary time flight for a projectile.
Projectile 4 - I can interpret/draw the position vs time graphs for a projectiles vertical and horizontal motion
Projectile 5 - I can interpret/draw the velocity vs time graph for a projectile's vertical and horizontal motion.
Projectile 6 - I can use motion concepts to solve problems involving projectiles in motion.
I will be demonstrating
Projectile 1 - I can represent the initial and final velocities of a projectile as vectors.
Projectile 2 - I can calculate the x and y components of a projectile's initial velocity.
Projectile 3 - I can calculate the necessary time flight for a projectile.
Projectile 4 - I can interpret/draw the position vs time graphs for a projectiles vertical and horizontal motion
Projectile 5 - I can interpret/draw the velocity vs time graph for a projectile's vertical and horizontal motion.
Projectile 6 - I can use motion concepts to solve problems involving projectiles in motion.
QuestionA soccer player kicks a stationary ball ,giving it a speed of 15m/s at an angle of 15degrees to the horizontal. (a) what is the maximum height reached by the ball? (b) what is the balls range? (c) how could the range be increased?
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AnswersStarting with the given information:
Displacement x = ? Displacement y = ? t = ? vx = ? voy = ? a = -9.8m/s^2 (gravity) I solved for Vx and Voy by creating a triangle of what the motion would receive at it's maximum height. Using the component equations V(sinΘ) and V(cosΘ) I was able to determine the vx and vy components. 15(sin15)=3.9m/s = Voy 15(cos15)= 14.49m/s = Vx |
I was able to determine Vy because as shown in the graph, the velocity will reach vy = 0 at it's highest peak going to the right.
In order to recieve maximum height (positive) we would need to solve for time. With the given information solved so far I am only able to use vy = at+voy. So using basic algebra, I plugged in my values and the time equals .4 seconds. Vy=at+voy 0=(-9.8m/s^2)t+3.9 (-3.9/9.8) = T T=.4 seconds. Now that I successfully solved for time, we are now able to solve for the maximum height using the position equation. Y =1/2(a)(t^2) + (vy)(t)+Vy and by putting in all my values, I was able to recieve .0776m as a displacement of height. Y=1/2(-9.8)(.4^2)+(3.9)(.4) + 0 Y = -.0784 + 1.56 Y= .776m Now for problem B. Since we solved for time it will be easy to solve. Time must be multiplied because if it takes .4 seconds to reach tis maximum height. Then it takes .8 seconds to find it's displacement. multiply that with vx and you get x(displacement) = 11.592. x(displacement) = (.8)(14.49) x(displacement) = 11.592m To increase the range of the ball, you could increase the angle or initial velocity and your values will change dramatically. Final information. Displacement x = 11.592m Displacement y = .776m t = .4 (displacement y) / .8 (displacement x) vx = 14.49m/s voy = 0m/s a = -9.8m/s^2 (gravity) |
In the xt graph, the position constantly goes up and never stops when the object is being kicked. The yt graph shows that the object goes up and goes back down due to gravity. Vx shows that it is constantly positive when the ball is traveling to the right. Vy shows that the ball has a decreasing velocity going up in the air and then increasing velocity descending. The motion graph is showed to the far right, it shows that the car is traveling to the right decreasing its velocity and for the time that it is in the air, it stops. The object then continues and making larger arrows and increasing velocity.