 Freefall: Projectiles Released at an Angle (2D-Motion) Printer Friendly Version
When a projectile is launched with a non-zero horizontal velocity, its trajectory takes on the shape of a parabola instead of just the linear trajectory it had when released either from rest or thrown straight up or down. There are now two dimensions to the motion which act independently of each other - that is, neither the projectile's horizontal velocity nor its horizontal position impact its instantaneous vertical velocity or position.  Before we work any mathematical problems involving projectiles released at an angle, let's first review the graphical behavior of objects exhibiting a uniform, negative acceleration and objects moving at a uniform speed.

A Graphical Review of Vertical Properties

Vertically, gravity will still accelerate the projectile at -9.8 m/sec2.

 losing speed while traveling in a positive direction position-times-t velocity-timev-t acceleration-timea-t   gaining speed while traveling in a negative direction position-times-t velocity-timev-t acceleration-timea-t   A Graphical Review of Horizontal Properties

But, horizontally, there is no acceleration since gravity acts at right angles to that velocity's component and there is no air resistance. Note in the original diagram shown at the top of the page that the horizontal spacing is uniform, that is, the projectile travels the same distance forward each second.

 traveling at a constant speed in a positive direction position-times-t velocity-timev-t acceleration-timea-t   Projectiles Released at an Angle

Consequently, when a projectile is released at an angle with a given speed its trajectory must be analyzed in two parts. Play the following physlet from Davidson College showing a ball thrown at an angle from the ground and note how these two simultaneous, yet independent, behaviors work together. In solving these problems we can always start with the following conditions: Horizontal Vertical timerelease speedrelease angle a = 0 a = - 9.8 m/sec2 vH = v cos(θ) vv = vo = v sin(θ) R = vHt s = vot + ½at2

In the table shown above, the variable, R, represents the range of the projectile; that is, the horizontal distance that the projectile travels from the point of release until it strikes the ground. Since there is no horizontal acceleration, the formula used to calculate a projectile's range, R = vHt , is derived from the equation d = rt for constant speed in which we substitute in range for distance, vH for rate, and leave time the same.

Notice, that the horizontal component vH is not renamed since it remains constant throughout the projectile's entire trajectory while vv is renamed as vo since the vertical velocity is constantly changing and vv is just the vertical component of its initial speed as it begins its trajectory.

Note that the time, t, the initial release speed, v, and the trajectory's angle, θ, cross between the columns since they are parameters that apply to the bahaviors in both columns; that is, they are common quantities.

You might want to take a moment to review your kinematics equations for uniformly accelerated motion prior to completing the next examples. To solve these problems we will use an H | V chart modeling the one shown above. This type of chart provides us a means of organizing our information and helps us apply the correct formulas for each component's behavior.

Refer to the following information for the next four questions.

In the following physlet from Boston University three projectiles are launched from the same level on flat ground. As shown in the picture below, each projectile reaches the same maximum height before returning to the ground, but the blue projectile has the largest range, followed by the green projectile. The red projectile has the smallest range. Rank the three from largest to smallest according to their time in flight. (a) blue > green > red    (b) red > green > blue    (c) red = green = blue

 Rank the three from largest to smallest according to their initial vertical component of velocity. (a) blue > green > red    (b) red > green > blue    (c) red = green = blue

 Rank the three from largest to smallest according to their initial horizontal component of velocity. (a) blue > green > red    (b) red > green > blue    (c) red = green = blue

 Rank the three from largest to smallest according to their initial speed. (a) blue > green > red    (b) red > green > blue    (c) red = green = blue

Refer to the following information for the next five questions.

A ball was kicked at a speed of 10 m/sec at a 37º and eventually returns to ground level further downrange.
 What were the horizontal and vertical components of the ball's velocity?

 How much time did it spend in the air?

 How far downrange did it land?

 How high did it rise in the air?

 Use the following physlet from the University of Virginia to determine at what other angle the ball could have been kicked to achieve the same range?

Refer to the following information for the next three questions.

In the remaining 2 seconds of a basketball game, a player makes a last ditch effort to save the game and complete a 3-point shot. The player releases the ball at the same height as the rim of the basket at a speed of 10 m/sec and an angle of 75º. He hopes that the ball will use up the remaining time on the clock as it sails into the hoop so that the other team will not have another opportunity to score. Determine if the horizontal distance from which the ball was released exceeds the 6-meter radius required to score 3 points.
 What were the horizontal and vertical components of the ball's release velocity?

 How much time did it spend in the air?

 How far away from the basket was the player when he released the ball?

Refer to the following information for the next three questions.

In a baseball game, a batter strikes a ball at an angle of 45º in such a manner that it leaves the bat at the same height as the outfield fence 90 meters away. If the ball just clears the fence so that the play is a homerun, how fast did the ball leave the bat?
 What were the horizontal and vertical components of the ball's release velocity?

 Write an expression for the ball's range in terms of the variables v and t.

 Write an expression for the ball's vertical displacement in terms of the variables v and t and then solve for v.

Refer to the following information for the next five questions.

While standing on a 30 meter bridge, a fisherman tosses some unused bait off the bridge at a speed of 10 m/sec at an angle of  37º.
 What were the horizontal and vertical components of the ball's velocity?

 How much time did it spend in the air?

 How far downrange does it land in the water from the base of the bridge?

 At what speed did it enter the water?

 Use the following physlet from Simulations for High School Physics to determine the angle at which the bait could have been released to achieve maximum range.

To see an application of this lesson, reference the Monkey and the Hunter Problem and its accompanying video lab. Related Documents