Question 11.SP.10: A projectile is fired from the edge of a 150-m cliff with an...
A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle of 30° with the horizontal. Neglecting air resis-tance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground, (b) the greatest elevation above the ground reached by the projectile.
STRATEGY: This is a projectile motion problem, so you can consider the vertical and horizontal motions separately. First determine the equations governing each direction, and then use them to find the distances.
MODELING and ANALYSIS: Model the projectile as a particle and neglect the effects of air resistance. The vertical motion has a constant acceleration. Choosing the positive sense of the y axis upward and placing the origin O at the gun (Fig. 1), you have
Substitute these values into the equations for motion with constant accel-eration. Thus,
\begin{array}{rlrl}v_{y} & =\left(v_{y}\right)_{0}+a t & v_{y} & =90-9.81 t (1) \\y & =\left(v_{y}\right)_{0} t+\frac{1}{2} a t^{2} & y & =90 t-4.90 t^{2} (2)\\v_{y}^{2} & =\left(v_{y}\right)_{0}^{2}+2 a y &v_{y}^{2} & =8100-19.62 y (3)\end{array}The horizontal motion has zero acceleration. Choose the positive sense of the x axis to the right (Fig. 2), which gives you
\left(v_{x}\right)_{0}=(180 \mathrm{~m} / \mathrm{s}) \cos30^{\circ}=+155.9 \mathrm{~m} / \mathrm{s}Substituting into the equation for constant acceleration, you obtain
x=\left(v_{x}\right)_{0} t \quad x=155.9 t (4)a. Horizontal Distance. When the projectile strikes the ground,
y = -150 m
Substituting this value into Eq. (2) for the vertical motion, you have
-150 = 90t – 4.90t² t² – 18.37t – 30.6 = 0 t = 19.91 s
Substituting t= 19.91 s into Eq. (4) for the horizontal motion, you obtain
x = 155.9(19.91) x = 3100 m
b. Greatest Elevation. When the projectile reaches its greatest elevation,
v_{y} = 0; substituting this value into Eq. (3) for the vertical motion, you have
0= 8100 – 19.62y y= 413 m
Greatest elevation above ground = 150 m + 413 m = 563 m
REFLECT and THINK: Because there is no air resistance, you can treat the vertical and horizontal motions separately and can immediately write down the algebraic equations of motion. If you did want to include air resistance, you must know the acceleration as a function of the speed (you will see how to derive this in Chapter 12), and then you need to use the basic kinematic relationships, separate variables, and integrate.
