91Ó°ÊÓ

Given data:

The initial speed of the projectile is $v_o=65\frac{\mathrm{m}}{\mathrm{s}}$.

The angle of projection is $\mathrm{θ}=35°$.

The initial vertical position is $y_o=115\mathrm{m}$.

The final vertical position is $y=0$.

Assumptions:

Let the projectile take t time to reach point P.

The initial horizontal velocity of the projectile is $v_o\cos \mathrm{θ}$, and the initial vertical velocity of the projectile is $v_o\sin \mathrm{θ}$.

Part (a)

For the vertical motion,

$\begin{array}{c}y-y_o=\left(v_o\sin \mathrm{θ}\right)t-\frac{1}{2}g{t}^2\\ \left(\frac{g}{2}\right)t^2-\left(v_o\sin \mathrm{θ}\right)t+\left(y-y_o\right)=0\end{array}$

Now, calculating the value of t,

role="math" localid="1644921063282" $\begin{array}{rcl}t& =& \frac{-\left(-v_o\sin \mathrm{θ}\right)Â\pm \sqrt{{\left(-v_o\sin \mathrm{θ}\right)}^2-\left(4×\left(\frac{g}{2}\right)×\left(y-y_o\right)\right)}}{2×\left(\frac{g}{2}\right)}\\ & =& \frac{v_o\sin \mathrm{θ}Â\pm \sqrt{v_o^2{\sin }^2\mathrm{θ}-2g\left(y-y_o\right)}}{g}\\ & =& \frac{65\frac{\mathrm{m}}{\mathrm{s}}\sin {35.0}^oÂ\pm \sqrt{\left({\left(65\frac{\mathrm{m}}{\mathrm{s}}\right)}^2{\sin }^2{35.0}^o\right)-2\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(-115\mathrm{m}\right)}}{9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}}\\ & =& -2.37\mathrm{s},9.96\mathrm{s}\end{array}$

Here, you can find two values of the time but only $t=9.96\mathrm{s}$is acceptable.

Hence, the projectile hit point P after 9.96 s.

Part (b)

Now, the horizontal distance between the base of the cliff and the point P is

$\begin{array}{c}x=\left(v_o\cos \mathrm{θ}\right)t\\ =\left(65.0\frac{\mathrm{m}}{\mathrm{s}}×\cos 35°\right)×9.96\mathrm{s}\\ =530.32\mathrm{m}\end{array}$

Hence, the value of X is 530.32 m.

Part (c)

The horizontal component of its velocity is the same throughout the motion, and the horizontal component of the velocity is

$\begin{array}{c}{v}_x=v_o\cos \mathrm{θ}\\ =65.0\frac{\mathrm{m}}{\mathrm{s}}×\cos 35°\\ =53.24\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the horizontal component of the velocity at point P is $53.24\frac{\mathrm{m}}{\mathrm{s}}$.

Now, the vertical component of its velocity at point P is

$\begin{array}{c}{v}_y=v_o\sin {35}^o-gt\\ =\left(65.0\frac{\mathrm{m}}{\mathrm{s}}×\sin {35}^o\right)-\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}×9.96\mathrm{s}\right)\\ =-60.33\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the vertical component of the velocity at point P is $-60.33\frac{\mathrm{m}}{\mathrm{s}}$.

Part (d)

The magnitude of the velocity at point P is

$\begin{array}{c}v=\sqrt{v_x^2+v_y^2}\\ =\sqrt{{\left(53.24\frac{\mathrm{m}}{\mathrm{s}}\right)}^2+{\left(-60.33\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}\\ =80.46\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the magnitude of the velocity at point P is $80.46\frac{\mathrm{m}}{\mathrm{s}}$.

Part (e)

Let the velocity vector make $\mathrm{Ï•}$angle with the horizontal direction.

Then,

$\begin{array}{c}\tan \mathrm{ϕ}=\frac{v_{\mathrm{y}}}{v_{\mathrm{x}}}\\ \mathrm{ϕ}={\tan }^{-1}\left(\frac{-60.33\frac{\mathrm{m}}{\mathrm{s}}}{53.24\frac{\mathrm{m}}{\mathrm{s}}}\right)\\ \mathrm{ϕ}=-48.57°\end{array}$

Hence, the velocity vector makes a $48.57°$angle below the horizontal plane.

Part (f)

Let $y_{\max }$be the maximum height above the cliff.

The vertical velocity at the maximum height is $v_{y\max }=0$.

Then,

$\begin{array}{c}{v}_{y\max }^2={\left(v_o\sin 35°\right)}^2-2g{y}_{\max }\\ 0^2={\left(65\frac{\mathrm{m}}{\mathrm{s}}×\sin 35°\right)}^2-2\left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)y_{\max }\\ y_{\max }=70.92\mathrm{m}\end{array}$

Hence, the maximum height of the projectile above the cliff is 70.90 m.