Question

20 ft Determine the landing location (coordinates), impact speed and time of flight for the launch conditions shown. 4. 80 ft 20 ft

Determine the landing location (in coordinates), the impact speed, and the time of flight for the launch conditions shown.

Answer 1

Angle of projection is $\theta$.

From the given figure, $\cos\theta=\frac{3}{5},\sin\theta=\frac{4}{5}$ , $\cos\phi=\frac{2}{\sqrt{5}},\sin\phi=\frac{1}{\sqrt{5}}$

Initial velocity of projectile $u=20\,ft/s$

Horizontal distance traveled by projectile in time $t$ is  $x=u\cos\theta\, t$ ---(1)

Vertical distance traveled by projectile in time $t$ is $y=80+u\sin\theta t-\frac{1}{2}gt^2$ ---(2)

Coordinates of landing location are $x=20+l\cos\phi\,,$ ---(3)

$\,y=l\sin\phi$ ---(4)

from (1) and (3) ,$u\cos\theta \,t=20+l\cos\phi$

$20*\frac{3}{5} \,t=20+l*\frac{2}{\sqrt{5}}$$\Rightarrow l=\frac{(12 t-20)\sqrt{5}}{2}$---(5)

from (2) and (4)  $80+u\sin\theta t-\frac{1}{2}gt^2=l\sin\phi$

$20*\frac{4}{5} t-\frac{1}{2}*32.2*t^2=-80+l*\frac{1}{\sqrt{5}}$ ( since $g=32.2\,ft/s^2$ )

$l=(16 t-16.1*t^2+80)\sqrt{5}$ ---(6)

from (4) and (6) $(16 t-16.1*t^2+80)\sqrt{5}=\frac{(12 t-20)\sqrt{5}}{2}$

$\Rightarrow -16.1t^2+10 t+90=0$$\Rightarrow t=2.7\,s$

using $t=2.7\,s$ in equation (6) $l=(16 t-16.1*t^2+80)\sqrt{5}=13.04\,ft$

$x=20+l\cos\phi=20+13.04*\frac{2}{\sqrt{5}}=31.7\,ft$

$\,y=l\sin\phi=13.04*\frac{1}{\sqrt{5}}=5.8\,ft$

Final velocity components are $v_x=u\cos\theta$ and $v_y=u\sin\theta-gt$

$v_x=u\cos\theta=20*\frac{3}{5}=12\,ft/s$

$v_y=u\sin\theta-gt=20*\frac{4}{5}-(32.2)(2.7)=-70.94\,ft/s$

Magnitude of landing velocity (impact speed) $v=(v_x^2+v_y^2)^{1/2}=(12^2+70.94^2)^{1/2}=71.95\,ft/s$

Coordinates of landing location are $(x,y)=(31.7,5.8)\,ft$

Time of flight $t=2.7\,s$

impact speed $v=71.95\,ft/s$