This animation shows the parabolic path of a ball rolled off of something like a table and then allowed to fall freely. Click play to set it in motion. The projectile is the magenta ball; the green balls represent its horizontal and vertical velocities.
If a projectile released or hit at a 45° angle above horizontal should go th…
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The essence of how we treat projectile motion, the motion of a launched object after no more launching forces are working on it, is in this figure: The horizontal velocity (ignoring friction) is constant, and the vertical acceleration is just that of a freely-falling object.
That's the essence of projectile motion, no matter how complicated the scenario: Nature doesn't care about whether a projectile is moving horizontally. It's still going to be acted on by the force of gravity just as though it were dropped or thrown straight upward.
We'll need to break that initial velocity vector into its vertical and horizontal components to get the vertical velocity, vy and the horizontal velocity vx. To do that we can use the 45-45-90 triangle, which you should memorize.
This problem will require a little trigonometry because not all of those angles belong to convenient memorized triangles. The vertical (vy) and horizontal (vx) components of the muzzle velocity (that's just armament talk for the velocity of the cannonball or bullet as it exits the barrel) are shown here:
Let's look at another cannon problem. A cannon (muzzle velocity 110 m/s) located at the edge of a 50 m cliff is shot upward at a 55 angle. How far from the base of the cliff (assume that the cliff wall is vertical) will the cannonball land? Assume that the cannon muzzle is 1.5m above the flat ground on top of the cliff.
The previous diagrams, tables, and discussion pertain to how the horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory. Now we will investigate the manner in which the horizontal and vertical components of a projectile's displacement vary with time. As has already been discussed, the vertical displacement (denoted by the symbol y in the discussion below) of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity. Thus, the vertical displacement (y) of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion. This equation was discussed in Unit 1 of The Physics Classroom. The equation can be written as follows.
where g is -9.8 m/s/s and t is the time in seconds. The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance that a projectile falls if dropped from rest. It was also discussed earlier, that the force of gravity does not influence the horizontal motion of a projectile. The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally (vix) and the amount of time (t) that it has been moving horizontally. Thus, if the horizontal displacement (x) of a projectile were represented by an equation, then that equation would be written as
The diagram below shows the trajectory of a projectile (in red), the path of a projectile released from rest with no horizontal velocity (in blue) and the path of the same object when gravity is turned off (in green). The position of the object at 1-second intervals is shown. In this example, the initial horizontal velocity is 20 m/s and there is no initial vertical velocity (i.e., a case of a horizontally launched projectile).
Now consider displacement values for a projectile launched at an angle to the horizontal (i.e., a non-horizontally launched projectile). How will the presence of an initial vertical component of velocity affect the values for the displacement? The diagram below depicts the position of a projectile launched at an angle to the horizontal. The projectile still falls 4.9 m, 19.6 m, 44.1 m, and 78.4 m below the straight-line, gravity-free path. These distances are indicated on the diagram below.
where viy is the initial vertical velocity in m/s, t is the time in seconds, and g = -9.8 m/s/s (an approximate value of the acceleration of gravity). If a projectile is launched with an initial vertical velocity of 19.6 m/s and an initial horizontal velocity of 33.9 m/s, then the x- and y- displacements of the projectile can be calculated using the equations above. A sample calculation is shown below.
The data in the table above show the symmetrical nature of a projectile's trajectory. The vertical displacement of a projectile t seconds before reaching the peak is the same as the vertical displacement of a projectile t seconds after reaching the peak. For example, the projectile reaches its peak at a time of 2 seconds; the vertical displacement is the same at 1 second (1 s before reaching the peak) is the same as it is at 3 seconds (1 s after reaching its peak). Furthermore, the time to reach the peak (2 seconds) is the same as the time to fall from its peak (2 seconds).
4. The diagram below shows the trajectory for a projectile launched non-horizontally from an elevated position on top of a cliff. The initial horizontal and vertical components of the velocity are 8 m/s and 19.6 m/s respectively. Positions of the object at 1-second intervals are shown. Determine the horizontal and vertical velocities at each instant shown in the diagram.
so the velocity at 1.50 seconds is 61.9743 m/s x + 27.8937 m/s y (in vector component notation) It has a magnitude of (61.97432 + 27.89372)1/2 = 67.96 = 68.0 m/s At an angle of tan-1(27.8937/61.9743) = 24.2o above the horizontal (d)
38. A rescue plane wants to drop supplies to isolatedmountain climbers on a rocky ridge 325 m below. If the plane is travelinghorizontally with a speed of 250 km/h (69.4 m/s), (a) how far in advance of therecipients (horizontal distance) must the goods be dropped (Fig. 3-41 a) ? (b)Suppose, instead that the plane releases the supplies a horizontal distance of425 m in advance of the mountain climbers. What vertical velocity (up or down)should the supplies be given so that they arrive precisely at the climbers'position (Fig. 3-41 b) ? (c) With what speed do the supplies land in the lattercase?
58. William Tell must split the apple atop his son's headfrom a distance of 27 m. When he aims directly at the apple, the arrow ishorizontal. At what angle must he aim it to hit the apple if the arrow travelsat a speed of 35 m/s?
69. The cliff divers of Acapulco push off horizontallyfrom rock platforms about 35 m above the water, but they must clear rockyoutcrops at a water level that extend out of the water 5.0 m from the base ofthe cliff directly under their launch point. See Fig. 3-49. What minimum pushoffspeed is necessary to do this? How long are they in the air?
71. Agent Tim, flying a constant 185 km/h horizontallyin a low-flying helicopter, wants to drop a small explosive onto a mastercriminal's automobile traveling 145 km/h on a level highway 88.0 m below. Atwhat angle (with the horizontal) should the car be in his sights when the bombis released (Fig. 3-51)?
The magnitude is: ( 25.6579 m/s2 + 57.6347 m/s2)1/2 = 63.0880 m/s = 63 m/s (magnitude) The angle is tan-1(57.6347/25.6579) = 66.0023o = 66o degrees above horizontal. (Table of contents)
During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of[latex]\boldsymbol75.0^o[/latex]above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?
Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which[latex]\boldsymbola_x=0[/latex]and[latex]\boldsymbola_y=-g.[/latex]We can then define[latex]\boldsymbolx_0[/latex]and[latex]\boldsymboly_0[/latex]to be zero and solve for the desired quantities.
24: Prove that the trajectory of a projectile is parabolic, having the form[latex]\boldsymboly=ax+bx^2.[/latex]To obtain this expression, solve the equation[latex]\boldsymbolx=v_0xt[/latex]for[latex]\boldsymbolt[/latex]and substitute it into the expression for[latex]\boldsymboly=v_0yt-(1/2)gt^2[/latex](These equations describe the[latex]\boldsymbolx[/latex]and[latex]\boldsymboly[/latex]positions of a projectile that starts at the origin.) You should obtain an equation of the form[latex]\boldsymboly=ax+bx^2[/latex]where[latex]\boldsymbola[/latex]and[latex]\boldsymbolb[/latex]are constants.
We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range
A plane is traveling from Portland to Seattle, which is 100 miles due north of Portland. There is a constant wind traveling southeast at 30 mph. If the plane needs to get to Seattle in one hour while flying due north, at what speed (relative to the wind) and angle should the pilot fly?
Suppose that a golf ball is struck such that it travels at a speed of at an angle to the horizontal. Neglecting air resistance, how long will the golf ball remain in the air before it touches the ground again?
In a soccer kick the filtered velocity data do not produce accurate measures of the ball projection velocity because of the rapid change in the velocity of the ball during the foot-ball impact (Knudson and Bahamonde, 2001). Instead, the ball projection velocity was calculated using unfiltered ball displacement data from images immediately after the instant of projection. The horizontal component of the ball velocity was calculated as the first derivative of a linear regression line fitted to the unfiltered ball displacement data, and the vertical component of the ball velocity was calculated as the first derivative of a quadratic regression line (with the second derivative set equal to -9.81 ms-2) fitted to the unfiltered ball displacement data (Nunome et al., 2006). Eight images were used in the calculation of the ball velocity components as this gave the best compromise between the decrease in uncertainty and the decrease in accuracy (due to the increasing effect of aerodynamic drag) as the number of data points was increased. The projection velocity and projection angle of the ball were calculated from the horizontal and vertical velocities of the ball. The projection height was the vertical distance of the center of mass of the ball relative to the ground at the instant of projection, and the projection distance was the horizontal distance of the center of mass of the ball relative to the kicking line at the instant of projection. The effective kick distance was calculated by adding the projection distance to the measured kick distance. The spin rate of the ball was calculated from the rotation of the ball markings (three hoops) over 5-20 frames after impact. As this was a 2-D study, the ball spin was only calculated for a spin direction that was perpendicular to the plane of the kick (i.e., backspin or topspin). 2ff7e9595c
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