Sample Problem 11-2: Firing angle of a projectile

A projectile is fired at a speed of 210 m/s and an angle θ. The projectile's intended target is 2,600 m away and 350 m above the firing point.

Solution

(a) The motion of the projectile can be analyzed by considering the horizontal and vertical components. The initial velocity v₀ can be resolved into orizontal and vertical components:

v_0x = v₀cos(θ) and v_0y = v₀sin(θ)

In the horizontal direction the velocity is constant, and the position of the projectile as a function of time is given by:

x = v_0xt

Substituting x = 2600 m for the horizontal distance that the projectile travels to reach the target and 210cos cos(θ) for v_0x, and solving for t gives:

In the vertical direction the position of the projectile is given by:

Substituting y = 350 m for the vertical coordinate of the target, 210sin(θ) for v_0x, g = 9.81, and t gives:

The solution of this equation gives the angle θ at which the projectile has to be fired.

(b) A solution of the equation derived in part (a) obtained by using the solve command (in the Command Window) is:

(c) The solution from part (b) shows that there are two possible angles and thus two trajectories. In order to make a plot of a trajectory, the x and y coordinates of the projectile are written in terms of t (parametric form):

The domain for t is t = 0 to .

These equations can be used in the ezplot command to make the plots shown in the following program written in a script file.

When this program is executed, the following plot is generated in the Figure Window: