Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. write down the general equations of motion for the object. which condition needs to be imposed to maximize the total distance traveled as a function of the launch angle for fixed launch speed vo? solve the equation and show that the optimal launch angle is 45° as a numerical example, calculate the distance traveled as a function of launch angle for vo m/s and 0 -0, 15°, 30°, 45°, 60, 75o, and 90°. graph your results for part (d) of the problem, with the launch angle along the horizontal axis and the horizontal range of the projectile along the vertical axis. a. b. c. d. e. answers: () show that! (d) 0, 0.051 m, 0.088 m , 0.102 m, 0.088 m, 0.051 m, 0

sin  2θ = 1    θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

            R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

          R = vo² Sin (2 15) /9.8

          R = Vo² 0.051 m

In the table are all values in two ways

Angle (θ)                  distance R (x)

 0                 0                     0

15                 0.051 Vo²        0.5 Vo²/g

30                0.088 vo²        0.866   Vo²/g

45                0.102 Vo²        1   Vo²/g

60                0.088 Vo²      0.866   Vo²/g

75                0.051 vo²        0.5   Vo²/g

90                0                     0

See graphic ( R Vs θ)  in the attached ¡, it can be done with any program, for example EXCEL

the ima (higher than 1m) will be greater than the ima with the height of 1m.