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 ${\vec {a}}=-9.80{\textrm {m/s}}^{2}{\hat {j}}<math>.<math>{\vec {v}}={\vec {v}}_{0}+{\vec {a}}t\rightarrow {\vec {v}}_{0}={\vec {v}}-{\vec {a}}t$

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

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