Physics: Two-Dimensional Projectile Motion
Resolving vectors and predicting projectile paths
Physics: Two-Dimensional Projectile Motion
Resolving vectors and predicting projectile paths
Physics - Grade 9-12
- 1
A soccer ball is kicked with an initial speed of 25 m/s at an angle of 40 degrees above the horizontal. Find the horizontal and vertical components of the initial velocity.
Use cosine for the component adjacent to the angle and sine for the component opposite the angle.
The horizontal component is vx = 25 cos(40 degrees) = 19.2 m/s. The vertical component is vy = 25 sin(40 degrees) = 16.1 m/s. - 2
A ball rolls horizontally off a 45 m tall cliff with a horizontal speed of 20 m/s. How long does it take to hit the ground, and how far from the base of the cliff does it land?
The time comes from vertical motion: t = sqrt(2h/g) = sqrt(90/9.8) = 3.03 s. The horizontal distance is x = vt = 20 times 3.03 = 60.6 m, so the ball lands about 60.6 m from the base. - 3
A projectile is launched from level ground at 18 m/s and 30 degrees above the horizontal. Find its total time in the air and its maximum height.
For a projectile that lands at the same height, the upward time equals the downward time.
The initial vertical velocity is vy = 18 sin(30 degrees) = 9.0 m/s. The total time is t = 2vy/g = 2(9.0)/9.8 = 1.84 s. The maximum height is h = vy^2/(2g) = 81/(19.6) = 4.13 m. - 4
Explain why the horizontal velocity of a projectile stays constant while the vertical velocity changes during ideal projectile motion.
The horizontal velocity stays constant because there is no horizontal acceleration when air resistance is ignored. The vertical velocity changes because gravity provides a constant downward acceleration of 9.8 m/s^2. - 5
A football is kicked from level ground at 22 m/s and 35 degrees above the horizontal. Find the range of the football if it lands at the same height it was kicked.
Use the range equation only because the projectile starts and lands at the same height.
The range is R = v^2 sin(2theta)/g. Substituting gives R = 22^2 sin(70 degrees)/9.8 = 46.4 m, so the football travels about 46.4 m horizontally. - 6
A projectile has an initial horizontal velocity of 12 m/s and an initial vertical velocity of 16 m/s. Find its initial speed and launch angle above the horizontal.
The initial speed is v = sqrt(12^2 + 16^2) = 20 m/s. The launch angle is theta = tan^-1(16/12) = 53.1 degrees above the horizontal. - 7
Two balls are released from the same height at the same time. Ball A is dropped straight down. Ball B is fired horizontally. Which ball hits the ground first, and why?
Compare only the vertical motion of the two balls.
The balls hit the ground at the same time because they have the same initial vertical velocity and the same vertical acceleration due to gravity. Ball B also moves horizontally, but that does not affect its fall time. - 8
A projectile is launched from the ground at 30 m/s and 60 degrees above the horizontal. Find its horizontal position and height after 2.0 s.
The horizontal velocity is vx = 30 cos(60 degrees) = 15 m/s, so x = 15(2.0) = 30 m. The initial vertical velocity is vy = 30 sin(60 degrees) = 26.0 m/s, so y = 26.0(2.0) - 4.9(2.0)^2 = 32.4 m. - 9
A cannonball reaches its maximum height 3.5 s after launch. Find the initial vertical velocity and the maximum height above the launch point.
At the top of the path, the vertical velocity is momentarily zero.
At maximum height, the vertical velocity is 0 m/s. The initial vertical velocity is vy = gt = 9.8(3.5) = 34.3 m/s. The maximum height is h = vy^2/(2g) = 34.3^2/19.6 = 60.0 m. - 10
A ball is kicked at 16 m/s from level ground and lands 20 m away at the same height. Find the two possible launch angles that could produce this range.
For projectiles that start and land at the same height, complementary angles can give the same range.
Using R = v^2 sin(2theta)/g gives sin(2theta) = Rg/v^2 = 20(9.8)/16^2 = 0.766. Thus 2theta is about 50.0 degrees or 130.0 degrees, so theta is about 25.0 degrees or 65.0 degrees. - 11
A marble rolls horizontally off a table that is 0.90 m high and lands 1.40 m from the table edge. What was the marble's horizontal speed as it left the table?
The fall time is t = sqrt(2h/g) = sqrt(1.80/9.8) = 0.429 s. The horizontal speed is vx = x/t = 1.40/0.429 = 3.26 m/s. - 12
Describe the shapes of these graphs for ideal projectile motion: horizontal position vs. time, vertical position vs. time, horizontal velocity vs. time, and vertical velocity vs. time.
Think about which quantities are constant and which have constant acceleration.
Horizontal position vs. time is a straight line because horizontal velocity is constant. Vertical position vs. time is a parabola that curves downward. Horizontal velocity vs. time is a horizontal line. Vertical velocity vs. time is a straight line with a negative slope because gravity reduces upward velocity and increases downward velocity. - 13
An airplane flying horizontally at 70 m/s drops a package from an altitude of 490 m. How long does the package take to reach the ground, and how far horizontally does it travel before landing?
The package keeps the airplane's horizontal velocity at the moment it is released.
The time to fall is t = sqrt(2h/g) = sqrt(980/9.8) = 10.0 s. The horizontal distance is x = vt = 70(10.0) = 700 m. - 14
A projectile is launched from the ground at 10 m/s and 45 degrees above the horizontal. Find its displacement components and speed after 1.2 s.
The initial components are vx = 7.07 m/s and vy = 7.07 m/s. After 1.2 s, the displacement components are x = 7.07(1.2) = 8.49 m and y = 7.07(1.2) - 4.9(1.2)^2 = 1.43 m. The velocity components are vx = 7.07 m/s and vy = 7.07 - 9.8(1.2) = -4.69 m/s, so the speed is sqrt(7.07^2 + (-4.69)^2) = 8.48 m/s. - 15
For a projectile launched and landing at the same height with no air resistance, what launch angle gives the greatest range? Explain how the answer changes if the landing point is much lower than the launch point.
A lower landing point gives the projectile more flight time even with a flatter launch.
When the launch and landing heights are the same, a 45 degree launch angle gives the greatest range. If the landing point is much lower than the launch point, the best angle is usually less than 45 degrees because the projectile has extra time to fall while moving horizontally.