Concepts A
Concepts B
Concepts C
Problems A
Problems B
100
State all four measurable variables used in kinematics AND their units of measurement.
Displacement (m)
Velocity (m/s)
Acceleration (m/s2)
Time (s)
100
A bowling ball and a tennis ball are dropped from the same height. Which hits the ground first (in less time)?
Same time!
(g is constant on all objects!)
100
Does the speedometer of a car measure speed, velocity, or both?
Speedometers only measure speed. They do not indicate direction.
100
A golf ball is hit an an angle and lands 200m downrange (horizontal displacement) after 5 seconds of time. After how much time from the strike was the golf ball at its apex height?
2.5 seconds; time to apex if half of the total time of flight
100
A rifle with a 1 m long barrel can fire a bullet with a velocity of 400 m/s. Calculate the acceleration of the bullet during the firing process.
vf2 = vi2 + 2ad
(400m/s)2 = 02 + (2)a(1m)
a = 80,000 m/s2
200
Describe the acceleration that acts on projectiles (include the physical reason it exists and its magnitude and direction).
Acceleration on projectiles is due to gravity.
It acts downward exclusively at 9.81m/s2
200
Two identical tennis balls are dropped from the same height. Ball A is launched horizontally while ball B is dropped simultaneously. Which ball hits the ground first?
Same time!
(g is constant on all objects, and horizontal motion is completely independent from vertical motion)
200
An airplane is flying from New York to California, 5 million meters (5 x 106 m) away. This trip takes approximately 6.313 hours. What is the acceleration of this plane during the flight if it was moving at a constant velocity of 220 m/s?
0 m/s2
Constant velocity!
200
A golf ball is hit an an angle and lands 200m downrange (horizontal displacement) after 5 seconds of time. What was the ball’s horizontal initial velocity component?
dx = vixt
200m = vix(5sec)
vix = 40 m/s
200
A rifle is held horizontally at 1.5 m above the ground. It fires a bullet with a velocity of 400 m/s. The bullet hits the ground some time later. What is the bullet’s time of flight?
t = √[(2dy) / g]
t = √[(2(1.5m)) / 9.81m/s2]
t = 0.553 sec
300
What is the vertical velocity of projectiles at their apex height?
0 m/s
This is the transition point between upward velocity (+) and downward velocity (-)
300
A projectile is fired at an angle of 50°, while another is fired at 45° at the same initial velocity. Which projectile has a greater time of flight? (Must explain why)
50° launch has greater time; higher launch angles result in greater vertical velocity, time, and maximum height at the expense of horizontal velocity and horizontal displacement
300
Is it possible for a projectile to have velocity upward but acceleration downward? Explain.
Yes. The projectile is decelerating. This occurs in the vertical direction while the projectile is rising.
300
A golf ball is hit an an angle and lands 200m downrange (horizontal displacement) after 5 seconds of time. What was the ball’s vertical initial velocity component?
Using vertical info to apex
vfy = viy + ayt
0m/s = viy + (-9.81m/s2)(2.5s)
viy = 24.525 m/s
300
A rifle is held horizontally at 1.5 m above the ground. It fires a bullet with a velocity of 400 m/s. The bullet hits the ground some time later. What is the bullet’s horizontal displacement?
dx = vxt
dx = (400m/s)(0.553sec)
dx = 221.2 m
400
Draw a vector to represent the velocity of a projectile at the apex point of its trajectory.
→
Velocity only acts horizontally at apex (vy = 0)
400
A projectile is fired at an angle of 50°, while another is fired at 45° at the same initial velocity. Which projectile has a greater horizontal displacement? (Must explain why)
45° launch; this is the optimal angle for maximum horizontal displacement because it results in the maximum combination (product) between horizontal velocity and time
400
In which direction is a projectile's acceleration at the point indicated in this picture? (The point is before the projectile reaches its apex height.)
Downward
Acceleration on projectiles is ALWAYS downward at all points a constant rate of 9.81m/s2
400
A golf ball is hit an an angle and lands 200m downrange (horizontal displacement) after 5 seconds of time. What was the maximum vertical displacement (apex or max. altitude) of the ball during its flight?
vfy2 = viy2 + 2aydy
(0m/s)2 = (24.525m/s)2 + 2(-9.81m/s2)(dy)
dy = 30.66 m
400
A baseball is thrown at 25 m/s at an angle of 30° above the horizontal. Solve for both the baseball’s horizontal and vertical velocity components.
vix = vicos(𝜃)
vix = (25m/s)cos(30°)
vix = 21.65 m/s
viy = visin(𝜃)
viy = (25m/s)sin(30°)
viy = 12.5 m/s
500
Compare a projectile’s initial launch velocity to its final impact velocity, assuming a symmetrical trajectory. (please reference velocity components).
Horizontal velocity is constant throughout the projectiles flight. Vertical velocity is reversed in direction but equal in magnitude between launch and impact
500
A projectile is fired at an angle of 25°, while another is fired at 65° at the same initial velocity. Compare and explain these projectiles’ horizontal displacements (ranges) and maximum vertical displacements (heights).
Equal horizontal displacements (angles are the same difference from 45); higher angle means higher maximum height and time of flight, but the combination of vx and time (dx) is equal for both
500
Is it possible for an object to have 0 velocity, but have a non-zero acceleration at the same time? Explain.
Yes. Just because an object isn’t moving at one moment (0 velocity) does not mean it won’t begin moving (due to its acceleration) in the next moment. This occurs at projectiles’ max. height when considering their vertical motion, or in a freefall scenario where vi =0m/s but the object begins to move in the next moment due to gravitational acceleration.
500
A golf ball is hit an an angle and lands 200m downrange (horizontal displacement) after 5 seconds of time. At what velocity and angle was the ball initially launched?
vi = √[(vix)2 + (viy)2]
vi =√[(40m/s)2+(24.525m/s)2]
vi = 46.920 m/s
𝜃 = tan-1(viy / vix)
𝜃 = tan-1(24.525m/s / 40m/s)
𝜃 = 31.51°
500
A baseball is thrown at 25 m/s at an angle of 30° above the horizontal. Solve for the baseball’s horizontal displacement (upon reaching its initial vertical position on the way down).
Range = [vi2sin(2𝜃)] / g
Range = [(25m/s)2sin(60°)] / (9.81m/s2)
Range = 55.715 m