Answer to Question #257129 in Mechanics | Relativity for Anne

Question

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a. What is the maximum height reached by the object? b. What is the total flight time (between launch and touching the ground) of the object? c. What is the magnitude of the velocity of the object just before it hits the ground?

Answer 1

(a) Let's first find the time that the object takes to reach the maximum height:

$v_y = v_{0y}-gt,$ $0=v_0sin\theta-gt,$ $t=\dfrac{v_0sin\theta}{g}.$

We can find the maximum height reached by the object from the kinematic equation:

$y_{max}=v_0tsin\theta-\dfrac{1}{2}gt^2.$

Substituting $t$ into the second equation, we get:

$y_{max}=\dfrac{v_0^2sin^2\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2=\dfrac{v_0^2sin^2\theta}{2g},$ $y_{max}=\dfrac{(20\ \dfrac{m}{s})^2\times sin^225^{\circ}}{2\times9.8\ \dfrac{m}{s^2}}=3.64\ m.$

(b) We can find the total flight time as follows:

$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$ $t_{flight}=\dfrac{2\times20\ \dfrac{m}{s}\times sin25^{\circ}}{9.8\ \dfrac{m}{s^2}}=1.72\ s.$

(c) Let's first find $x$- and $y-$components of the object's velocity just before it hits the ground:

$v_x=v_0cos\theta=20\ \dfrac{m}{s}\times cos25^{\circ}=18.1\ \dfrac{m}{s},$ $v_y=v_{0y}-gt_{flight}=v_0sin\theta-g\dfrac{2v_0sin\theta}{g}=-v_0sin\theta,$ $v_y=-20\ \dfrac{m}{s}\times sin25^{\circ}=-8.45\ \dfrac{m}{s}.$

Finally, we can find the magnitude of the velocity of the object just before it hits the ground from the Pythagorean theorem:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(18.1\ \dfrac{m}{s})^2+(-8.45\ \dfrac{m}{s})^2}=20\ \dfrac{m}{s}.$