In Fig. 4-34, a stone is projected at a cliff of height h with an initial speed of${}^{\mathbf{42}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$directed at angle${}^{{\mathbf{\theta }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}}$above the horizontal. The stone strikes at${}^{\mathbf{A}}$,${}^{\mathbf{5}\mathbf{.}\mathbf{50}\mathbf{}\mathit{s}}$after launching. Find (a) the height${}^{\mathit{h}}$of the cliff, (b) the speed of the stone just before impact at${}^{\mathbf{A}}$, and (c) the maximum height${}^{\mathbf{H}}$reached above the ground.

$v_0=42.0m/s$

Projection of angle is${}^{{\theta }_0=60°}$

In this case${}^{y=h}$

This problem deals with the kinematic equations and projectile motion of an object. Kinematic equations are the equations that describe the motion of an object with constant acceleration. Projectile motion is a form of motion experienced by an object or particle that moves under the action of gravity only.

Applying second kinematic equation, the height of the cliff can be found. Using standard notation for the magnitude, the speed of the stone just before the impact will be calculated. Using the formula for the height of the projectile, find the maximum height reached by the stone above the ground.

Formula:

The displacement in kinematic equation can be written as,

$y=v_{0}sin{\theta }_{0}t-\frac{1}{2}g{t}^2$ (i)

Magnitude of the velocity is given by,

localid="1654579185417" $v=\sqrt{{\left(\stackrel{\rightharpoonup }{v_x}\right)}^2+{\left(\stackrel{\rightharpoonup }{v_y}\right)}^2}$ (ii)

$H=\frac{{\left(v_0\sin {\theta }_0\right)}^2}{2g}$ (iii)

Here${}^{y=h}$thus, equation (i) can be written as

$h=v_0\sin {\theta }_{0}t-\frac{1}{2}g{t}^2$ (iv)

$$\begin{gathered} {\mathrm{v}}_0=42.0\mathrm{m}/\mathrm{s} \\ {\mathrm{\theta }}_0=60.0° \\ \mathrm{t}=5.50\mathrm{s} \end{gathered}$$

With the above values, equation (iv) can be written as,

$h=42.0\times sin\left(60\right)\times 5.5-\frac{1}{2}\times 9.8\times 5.{50}^2$

Thus,

$h=51.8m$

Vertical component of velocity${}^{v_x=v_0\cos {\theta }_0}$

Horizontal component of velocity${}^{{\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_0{\mathrm{sin\theta }}_0-\mathrm{gt}}$

So, using equation (ii) the magnitude of the velocity will be,

$$\begin{gathered} \mathrm{v}=\sqrt{{\left(42.0\cos \left(60\right)\right)}^2+{\left(42.0\sin \left(60\right)-\mathrm{gt}\right)}^2} \\ \mathrm{v}=27.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (ii), the maximum height is given by,

$H=\frac{{\left(\begin{array}{c}42.0\sin \left(60\right)\end{array}\right)}^2}{2\times 9.8}$

Thus,

$H=67.5m$