Question

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.

Short answer

(a) Height${}^h$of the cliff is${}^{51.8m}$

(b) Speed of the stone just before impact at A is${}^{27.4m/s}$

(c) The maximum height H reached above the ground is${}^{67.5m}$

Step-by-step solution

Given information

$v_0=42.0m/s$

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

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

To understand the concept of kinematic equations

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)

(a) to find theheighth of the cliff

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,

$h=51.8m$

(b) to find the speed of the stone just before the impact at A

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,

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

(c) To find the maximum height H reached above the ground

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,

$H=67.5m$