Chapter 3 Problem 48

Problem

Exactly 3.0 s after a projectile is fired into the air from the ground,it is observed to have a velocity ${\vec {v}}=\left(8.6{\hat {i}}+4.8{\hat {j}}\right)$ m/s, where the $x$ axis is horizontal and the $y$ axis is positive upward. Determine

(a) the horizontal range of the projectile,

(b) its maximum height above the ground, and

(c) its speed and angle of motion just before it strikes the ground.

Solution

The constant acceleration is ${\vec {a}}=-9.80{\textrm {m/s}}^{2}{\hat {j}}$.

${\vec {v}}={\vec {v}}_{0}+{\vec {a}}t\rightarrow {\vec {v}}_{0}={\vec {v}}-{\vec {a}}t$

$=\left(8.6{\hat {i}}+4.8{\hat {j}}\right){\textrm {m/s}}-\left(-9.80{\textrm {m/s}}^{2}{\hat {j}}\right)\left(3.0{\textrm {s}}\right)=\left(8.6{\hat {i}}+34.2{\hat {j}}\right){\textrm {m/s}}$

The initial speed: $v_{0}={\sqrt {\left(8.6{\textrm {m/s}}\right)^{2}+\left(34.2{\textrm {m/s}}\right)^{2}}}=35.26{\textrm {m/s}}$

The launch direction: $\theta _{0}=\tan ^{-1}{\frac {34.2{\textrm {m/s}}}{8.6{\textrm {m/s}}}}=75.88^{\circ }$

(a)

$y=y_{0}+v_{y0}t+{\frac {1}{2}}a_{y}t^{2}$ $\rightarrow 0=0+v_{y0}t+{\frac {1}{2}}a_{y}t^{2}$

$0=t\left(v_{y0}-{\frac {1}{2}}gt\right)$

$t=0$ is the start of the motion, thus the answer we seek is $t={\frac {2v_{y0}}{g}}$

$R=v_{x0}t=v_{x0}{\frac {2v_{y0}}{g}}={\frac {2v_{0}^{2}\sin \theta _{0}\cos \theta _{0}}{g}}$

$R={\frac {v_{0}^{2}\sin 2\theta _{0}}{g}}={\frac {\left(5.26{\textrm {m/s}}\right)^{2}\left(\sin(151.76^{\circ })\right)}{9.80{\textrm {m/s}}^{2}}}=6.0\times 10^{1}{\textrm {m}}$

(b)

The vertical velocity at the maximum height is 0, and the initial height is 0.

$v_{y}^{2}=v_{y0}^{2}+2a_{y}\left(y_{\textrm {max}}-y_{0}\right)\rightarrow$

$y_{\textrm {max}}=y_{0}+{\frac {v_{y}^{2}-v_{y0}^{2}}{2a_{y}}}={\frac {-\left(34.2{\textrm {m/s}}\right)^{2}}{2\left(-9.80{\textrm {m/s}}^{2}\right)}}=59.68m\approx 6.0\times 10^{1}{\textrm {m}}$

(c)

The final speed just before the projectile hits the ground is the same as the initial speed, and the angle is the same as the launching angle, but below the horizontal.

$v_{\textrm {final}}=35.26{\textrm {m/s}}$

$\theta _{\textrm {final}}=-75.88^{\circ }$ (below the horizontal)