# PROJECTILE MOTION EQUATIONS

## Projectile Motion Definition

Projectile motions take place in two dimensions such as thrown tennis balls and kicked footballs. Projectile motion equations are independent one-dimensional equations in x and y directions.

The trajectory of a thrown tennis ball is shown in the following figure. When the ball is thrown with an angle α, the horizontal component of velocity in x direction is V0Cosα and vertical component in y direction is V0Sinα.  The ball reaches the highest point at point P and it returns to the ground at point S. Although the effect of air resistance is important, in many cases it can be ignored. In the following analysis, the effect of air resistance will be ignored.

## Projectile Motion Equations

The following projectile motion equations are used to solve projectile motion problems. In x direction, there is no acceleration while there is acceleration in y direction.  In order to use these equations, initial parameters such as initial positions (x0,y0) and initial velocities (V0x and V0y ) are needed.  x0 and y0 can be arbitrarily chosen as 0. The velocity in the x direction never changes and V0x always remains V0cos α. However, the velocity in the y direction is V0sinα at the beginning (at t=0) and that value will change because of the acceleration in y direction. The acceleration in y direction is going to be negative 9.8 m/s^2 (-g) which is gravity.

$${ x }_{ t }={ x }_{ 0 }+{ V }_{ { 0 }_{ x } }t$$  (Eq-1)

$${ V }_{ { x }_{ t } }={ V }_{ { 0 }_{ x } }$$  (Eq-2)

$${ y }_{ t }={ y }_{ 0 }+{ V }_{ { 0 }_{ y } }t+{ 1 }/{ 2 }{ a }_{ y }{ t }^{ 2 }$$   (Eq-3)

$${ V }_{ { y }_{ t } }={ V }_{ { 0 }_{ y } }+{ a }_{ y }t$$   (Eq-4)

## Shape of the Projectile Motion

The shape of the motion can be found with the following steps. x0 and y0 are chosen as 0 (initial location of the object is at (x,y) = (0,0)). Eq-1 and Eq-3 can be rearranged as follows.

$${ x }_{ t }={ V }_{ 0 }\cos { \alpha } t$$  (Eq-5)

$${ y }_{ t }={ V }_{ 0 }\sin { \alpha } t-{ 1 }/{ 2 }g{ t }^{ 2 }$$  (Eq-6)

Time t can be eliminated by substituting following equality obtained from Eq-5 into Eq-6.

$$t=x/({ V }_{ 0 }\cos { \alpha } )$$  (Eq-7)

The result is following second order equation and this is equation of a parabola.

$$y=\tan { \alpha } x-{ { 1 }/{ 2 }g }/{ { ({ V }_{ 0 } }\cos { \alpha } ) }^{ 2 }{ x }^{ 2 }$$  (Eq-8)

## Time to Reach Highest Point

The time that the object comes to its highest point P can be calculated by using Eq-4 and it can be found when Vy is 0. That is the moment that the velocity in the y direction becomes 0. At that moment, the object must be its highest point. The time can be found with the following projectile motion equations.

$$0={ V }_{ y }={ V }_{ 0 }\sin { \alpha } -gt\Rightarrow { t }_{ p }=\frac { { V }_{ 0 }\sin { \alpha } }{ g }$$ (Eq-9)

## Maximum Height of Projectile Motion

The highest point above the ground can be obtained by using Eq-3 for the time tp which is found with Eq-9. The highest point h which is y at time tp is found with following equation.

$${ y }_{ { t }_{ p } }=\frac { { \left( { V }_{ 0 }\sin { \alpha } \right) }^{ 2 } }{ g } -\frac { 1 }{ 2 } g\frac { { \left( { V }_{ 0 }\sin { \alpha } \right) }^{ 2 } }{ { g }^{ 2 } }$$ (Eq-10)

$$h=\frac { { \left( { V }_{ 0 }\sin { \alpha } \right) }^{ 2 } }{ 2g }$$ (Eq-11)

The interpretation of the equation is as follows:

• It’s reasonable that the highest point gets higher when the value of V0 is bigger. If the object is thrown up at a higher speed, it will, of course, reach a higher point.
• If the angle α is increased it’s reasonable that the object will get higher. The highest possible height that the object can reach is when the angle α is 90 degrees for a given velocity.
• If the object is thrown up on the moon with the same initial speed, it will go much higher so it makes sense that g is at downstairs of the equation.

## Projectile Motion Time to Hit the Ground

The time that the object hits the ground at point S can be calculated in two ways. First way is with Eq-3 for the time when yt is 0. There will be two solutions for Eq-3 where yt is 0. The first solution is the time at the beginning when time 0. The second solution is ts.

The shape of the projectile motion is parabola and it’s completely symmetric about the vertical line passing through the point P. So the time that it takes to go up from point O to P shall be same with the time to go down from P to S. This is the second solution and the result is given with the following equation.

$${ t }_{ s }={ 2t }_{ p }=\frac { { 2V }_{ 0 }\sin { \alpha } }{ g }$$ (Eq-12)

## Horizontal Distance (Range) in Projectile Motion

Horizontal distance OS can be calculated by using projectile motion equations 1 and 12.

$$OS=\frac { 2{ { V }_{ 0 } }^{ 2 }\cos { \alpha } \sin { \alpha } }{ g } =\frac { { { V }_{ 0 } }^{ 2 }\sin { 2\alpha } }{ g }$$ (Eq-13)

From Eq-13, it’s immediately seen that the largest range for a given velocity is obtained when the object is thrown 45 degrees wrt. ground. When 45 deg is substituted in the formula for the α, sin2α is equal to 1.