Answer to Question #146709 in Physics for Althea

Question

Suppose the body in Sample 20.0 is projected at 24.5 m/s, 30° above the horizontal. Find the a. time of flight, b. range, and c. Maximum height.

Answer 1

Let's write the equations of motion of the body in horizontal and vertical directions:

$x=v_0tcos\theta, (1)$ $y=v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$

here, $x$ is the horizontal displacement of the body (or the range), $v_0=24.5\ \dfrac{m}{s}$ is the initial velocity of the body, $t$ is the time of flight of the body, $\theta=30^{\circ}$ is the launch angle, $y$ is the vertical displacement of the body (or the height) and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

a) Let's first find the time that the body takes to reach the maximum height from the kinematic equation:

$v_y=v_0sin\theta-gt_{rise},$ $0=v_0sin\theta-gt_{rise},$ $t_{rise}=\dfrac{v_0sin\theta}{g}.$

Then, we can find the time of flight of the body:

$t=2t_{rise}=\dfrac{2v_0sin\theta}{g},$ $t=\dfrac{2\cdot 24.5\ \dfrac{m}{s}\cdot sin30^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.5\ s.$

b) We can substitute the time of flight of the body into the first equation and find the range of the body:

$x=24.5\ \dfrac{m}{s}\cdot 2.5\ s\cdot cos30^{\circ}=53\ m.$

c) Finally, we can substitute $t_{rise}$ into the second equation and find the maximum height:

$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$ $y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$ $y_{max}=\dfrac{(24.5\ \dfrac{m}{s})^2\cdot sin^230^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=7.66\ m.$

Answer:

a) $t=2.5\ s.$

b) $x=53\ m.$

c) $y_{max}=7.66\ m.$