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3.4 Projectile motion  (Page 2/16)

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Review of kinematic equations (constant a )

x = x 0 + v - t size 12{x=`x rSub { size 8{0} } `+` { bar {v}}t} {}
v - = v 0 + v 2 size 12{ { bar {v}}=` { {v rSub { size 8{0} } +v} over {2} } } {}
v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
v 2 = v 0 2 + 2 a ( x − x 0 ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a \( x - x rSub { size 8{0} } \) } {}
A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.
The total displacement s size 12{s} {} of a soccer ball at a point along its path. The vector s size 12{s} {} has components x size 12{x} {} and y size 12{y} {} along the horizontal and vertical axes. Its magnitude is s size 12{s} {} , and it makes an angle θ size 12{θ} {} with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used. The magnitude of the components of displacement s size 12{s} {} along these axes are x size 12{x} {} and y. size 12{y} {} The magnitudes of the components of the velocity v size 12{v} {} are v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {} and v y = v sin θ, size 12{v rSub { size 8{y} } =v"sin"θ} {} where v size 12{v} {} is the magnitude of the velocity and θ size 12{θ} {} is its direction, as shown in [link] . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Horizontal Motion ( a x = 0 ) size 12{"Horizontal Motion " \( a rSub { size 8{x} } =0 \) } {}
x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}
v x = v 0 x = v x = velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}
Vertical Motion ( assuming positive is up a y = − g = − 9. 80 m/s 2 ) size 12{"Vertical Motion " \( "assuming positive is up "a rSub { size 8{y} } = - g= - 9/"80"" m/s" rSup { size 8{2} } \) } {}
y = y 0 + 1 2 ( v 0 y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}
v y = v 0 y − gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}
y = y 0 + v 0 y t − 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}
v y 2 = v 0 y 2 − 2 g ( y − y 0 ) . size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) "."} {}

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t size 12{t} {} . The problem solving procedures here are the same as for one-dimensional kinematics    and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s size 12{s} {} and velocity v size 12{v} {} . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ = tan − 1 ( A y / A x ) size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} in the following form, where θ size 12{θ} {} is the direction of the displacement s size 12{s} {} and θ v size 12{θ rSub { size 8{v} } } {} is the direction of the velocity v size 12{v} {} :

Total displacement and velocity

s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}
θ = tan − 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } \( y/x \) } {}
v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}
θ v = tan − 1 ( v y / v x ) . size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) "."} {}
In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east.
(a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 size 12{a rSub { size 8{x} } =0} {} and v x size 12{v rSub { size 8{x} } } {} is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.

A fireworks projectile explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in [link] . 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?

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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