Question

1. you fire an angry bird with an initial velocity of 16 m/s at an angle of 27 degrees.

a. sketch a picture representing the problem.
b. what are the components of the initial velocity? fill in your x and y table
c. how long before the angry bird hits the ground?
d. what is the maximum height it reaches?
e. how far away does the angry bird land?

Explanation:

Step1: Find initial - velocity components

The initial velocity $v_0 = 16$ m/s and the angle $\theta=27^{\circ}$. The $x$ - component of the initial velocity is $v_{0x}=v_0\cos\theta$ and the $y$ - component is $v_{0y}=v_0\sin\theta$.
$v_{0x}=16\cos(27^{\circ})\approx16\times0.891 = 14.256$ m/s
$v_{0y}=16\sin(27^{\circ})\approx16\times0.454 = 7.264$ m/s

Step2: Find time of flight

The vertical - motion equation is $y = y_0+v_{0y}t-\frac{1}{2}gt^2$. When the bird hits the ground $y = y_0 = 0$. So, $0=v_{0y}t-\frac{1}{2}gt^2=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The other is $t=\frac{2v_{0y}}{g}$. Taking $g = 9.8$ m/s², $t=\frac{2\times7.264}{9.8}\approx1.482$ s

Step3: Find maximum height

At the maximum height, the vertical velocity $v_y = 0$. The equation $v_y^2=v_{0y}^2 - 2gh_{max}$ is used. Solving for $h_{max}$, we get $h_{max}=\frac{v_{0y}^2}{2g}$. Substituting $v_{0y}=7.264$ m/s and $g = 9.8$ m/s², $h_{max}=\frac{(7.264)^2}{2\times9.8}=\frac{52.765}{19.6}\approx2.692$ m

Step4: Find horizontal range

The horizontal motion is a uniform - motion with $x = v_{0x}t$. Substituting $v_{0x}=14.256$ m/s and $t = 1.482$ s, $x=14.256\times1.482\approx21.12$ m

Answer:

a. (Sketch: Draw a coordinate system with the origin as the launch - point. Draw a vector representing the initial velocity $v_0$ at an angle of $27^{\circ}$ with the horizontal. Label the magnitude of the vector as 16 m/s.)
b. $v_{0x}\approx14.26$ m/s, $v_{0y}\approx7.26$ m/s
c. $t\approx1.48$ s
d. $h_{max}\approx2.69$ m
e. $x\approx21.12$ m