# Projectile Motion Range and Maximum Height

#### Maverick56

An object launched at 30 degree angle (landing at same elevation) would have a maximum height that is 7 times its range.
Do we need to just memorize or can you show me mathematically?

#### ashtonjam

5+ Year Member
If the object had an initial launch speed of 10.0 m/s and was launched at 30 degrees to the horizontal,
Initial y velocity = 10sin(30) = 5 m/s
Initial x velocity = 10cos(30) = 8.66 m/s

Time it takes to get to the top of the flight: vfy = viy + at, where a = -9.81 m/s^2, viy = 5 m/s, and vfy = 0 m/s (Initial poin t = launch, Final point = top of path). You should find that t = 0.51 seconds.

The flight path is perfectly symmetrical, so the total time of flight is 0.51 seconds * 2 = 1.02 seconds.

Flight range is x velocity multiplied by time: Range = vx * t = 8.66 m/s * 1.02 s = 8.83 meters.

Max height can be calculated using 2ay = vfy^2 - viy^2, where a = -9.81 m/s^2, vfy = 0 m/s, and viy = 5 m/s (The initial point is the precise moment of launch and the final point is the top of the flight when vfy = 0 m/s). The calculated height is y = 1.27 meters.

So we found that range is 8.83 m and height is 1.27 m. 8.83/1.27 = 6.92 or so. If you didn't round, the answer becomes 7. This could be done using only variables as well, but having numbers to compare is nice.

#### SpaceBuff12

5+ Year Member
You have those numbers reversed. The horizontal distance traveled (range) is approximately 7 times the max height. I think it is silly to memorize as long as you understand how it is mathematically derived.

For example, an object launched 30 degrees above the horizon at 10 m/s will have an initial velocity vector of x: 10*cos(30) and y: 10*sin(30) (note: you should memorize the values of cosine and sine for angles 0, 30, 45, 60, and 90; this will greatly simplify some problems).

First look at the y-component to determine the time of flight for the object. At the apex of the trajectory the velocity in the y-direction is zero and thus we can solve:

v = v_y0 + a*t = 10*sin(30) -10t -> t = sin(30) = 0.5 seconds

It takes 0.5 seconds for the projectile to reach its apex and the total flight time is double (1 second).

The max height achieved is:

y = y_0 + v_y0*t +0.5*a*t^2 = 0 + 10*sin(30)*0.5 - 0.5*10*(0.5^2) = 1.25 m

In the x-direction:

x = x_0 + v_x0*t = 0 + 10*cos(30)*1 = ~8.66 m

8.66/1.25 ~ 7

#### SpaceBuff12

5+ Year Member
Ashtonjam beat me to it

#### ashtonjam

5+ Year Member
10 is a nice number after all!

5+ Year Member
Indeed.

#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
This is totally off topic, but I've lurked MCAT forums for about 3 years now, and literally, every person I've seen with the OP's avatar has done extremely well on the MCAT. Hope the trend continues