A BODY PROJECTED HORIZONTALLY
A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. Find the motorcycle’s position, distance from the edge of the cliff, and velocity 0.50 s after it leaves the edge of the cliff.
IDENTIFY and SET UP:
Figure 3.22 shows our sketch of the trajectory of motorcycle and rider. He is in projectile motion as soon as he leaves the edge of the cliff, which we take to be the origin (so x_{0}=y_{0}=0). His initial velocity \overrightarrow{\boldsymbol{v}}_{0} at the edge of the cliff is horizontal (that is, \alpha_{0}=0), so its components are v_{0 x}=v_{0} \cos \alpha_{0}=9.0 m/s and v_{0 y}=v_{0} \sin \alpha_{0}=0. To find the motorcycle’s position at t = 0.50 s, we use Eqs. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t) and (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}); we then find the distance from the origin using Eq. (3.23) (r=\sqrt{x^{2}+y^{2}}). Finally, we use Eqs.(3.21) (v_{x}=v_{0} \cos \alpha_{0}) and (3.22) (v_{y}=v_{0} \sin \alpha_{0}-g t to find the velocity components at t = 0.50 s.
EXECUTE:
From Eqs. (3.19) (x=\left(v_{0} \cos \alpha_{0}\right) t) and (3.20) (y=\left(v_{0} \sin \alpha_{0}\right) t-\frac{1}{2} g t^{2}), the motorcycle’s x- and y-coordinates at t = 0.50 s are
x=v_{0 x} t=(9.0 \mathrm{~m} / \mathrm{s})(0.50 \mathrm{~s})=4.5 \mathrm{~m}y=-\frac{1}{2} g t^{2}=-\frac{1}{2}\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(0.50 \mathrm{~s})^{2}=-1.2 \mathrm{~m}
The negative value of y shows that the motorcycle is below its starting point.
From Eq. (3.23) (r=\sqrt{x^{2}+y^{2}}), the motorcycle’s distance from the origin at t = 0.50 s is
From Eqs.(3.21) (v_{x}=v_{0} \cos \alpha_{0}) and (3.22) (v_{y}=v_{0} \sin \alpha_{0}-g t, the velocity components at t = 0.50 s are
v_{x}=v_{0 x}=9.0 \mathrm{~m} / \mathrm{s}v_{y}=-g t=\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(0.50 \mathrm{~s})=-4.9 \mathrm{~m} / \mathrm{s}
The motorcycle has the same horizontal velocity v_x as when it left the cliff at t = 0, but in addition there is a downward (negative) vertical velocity v_y. The velocity vector at t = 0.50 s is
\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath}=(9.0 \mathrm{~m} / \mathrm{s}) \hat{\imath}+(-4.9 \mathrm{~m} / \mathrm{s}) \hat{\jmath}From Eqs. (3.24) (v=\sqrt{v_{x}^{2}+v_{y}^{2}}) and (3.25) (\tan \alpha=\frac{v_{y}}{v_{x}}), at t = 0.50 s the velocity has magnitude v and angle a given by
v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(9.0 \mathrm{~m} / \mathrm{s})^{2}+(-4.9 \mathrm{~m} / \mathrm{s})^{2}}=10.2 \mathrm{~m} / \mathrm{s}\alpha=\arctan \frac{v_{y}}{v_{x}}=\arctan \left(\frac{-4.9 \mathrm{~m} / \mathrm{s}}{9.0 \mathrm{~m} / \mathrm{s}}\right)=-29^{\circ}
The motorcycle is moving at 10.2 m/s in a direction 29° below the horizontal.
EVALUATE: Just as in Fig. 3.17, the motorcycle’s horizontal motion is unchanged by gravity; the motorcycle continues to move horizontally at 9.0 m/s, covering 4.5 m in 0.50 s. The motorcycle initially has zero vertical velocity, so it falls vertically just like a body released from rest and descends a distance \frac{1}{2} g t^{2}=1.2 \mathrm{~m} in 0.50 s.
