Respuesta :
a. Horizontal range is given by,
[tex]Range=\frac{u^2sin2\theta}{g}[/tex]where u is the initial speed, and theta is the angle of projection,
Now velocity vector as a function of time is given by,
[tex]\vec{v}(t)=(ucos\theta)\hat{i}+(usin\theta-gt)\hat{j}[/tex]Given,
[tex]\begin{gathered} \vec{v}(3.3s)=8.6\text{ }\hat{i}+4.7\text{ }\hat{j} \\ \Rightarrow ucos\theta=8.6,usin\theta-(10)(3.3)=4.7 \\ \Rightarrow ucos\theta=8.6,usin\theta=37.7 \\ Dividing, \\ \Rightarrow tan\theta=\frac{37.7}{8.6}=4.3837 \\ \Rightarrow\theta=77.15\degree \\ Hence \\ ucos77.15\degree=8.6 \\ \Rightarrow u=\frac{8.6}{0.2224}=38.67\text{ m/s} \end{gathered}[/tex]And so, the horizontal range will be,
[tex]Range=\frac{(38.67)^2sin(154.3\degree)}{(10)}=64.85\text{ m}[/tex]b. Maximum height reached by a projectile is given by,
[tex]\begin{gathered} H_{max}=\frac{u^2sin^2\theta}{2g} \\ \Rightarrow H_{max}=\frac{(38.67)^2sin^2(77.15\degree)}{(2)(10)}=71.07\text{ m} \end{gathered}[/tex]c. Now time of flight of the projectile is given by,
[tex]t_{flight}=\frac{2usin\theta}{g}=\frac{(2)(38.67)sin(77.15\degree)}{(10)}=7.54\text{ s}[/tex]this is the time at which the projectile strikes the ground, and so its velocity then would be,
[tex]\begin{gathered} \vec{v}(7.54s)=8.6\text{ }\hat{i}+(37.7-10\times7.54)\hat{j} \\ \Rightarrow\vec{v}(7.54s)=8.6\text{ }\hat{i}-37.7\text{ }\hat{j} \end{gathered}[/tex]As expected the vertical component of velocity has just been flipped, and therefore its speed will be same as its initial speed i.e.
[tex]v_{strikes\text{ }ground}=u=38.67\text{ m/s}[/tex]d. Below the horizontal, let the angle be φ, then
[tex]tan\varphi=-\frac{v_y}{v_x}[/tex]extra -ve sign since its below the horizontal, when the projectile strikes the ground,
[tex]\begin{gathered} v_y=-37.7\text{ m/s} \\ v_x=8.6\text{ m/s} \\ \Rightarrow tan\varphi=-\frac{(-37.7)}{8.6}=\frac{37.7}{8.6}=tan\theta \\ \Rightarrow\varphi=\theta \\ \Rightarrow\varphi=77.15\degree \end{gathered}[/tex]Result: a. 64.85 m, b. 71.05 m, c. 38.67 m/s, d. 77.15°.
