Question

A Projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-56). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

FIGURE 3-56Problem 70

Short answer

The projectile is launched with a velocity of $60.50\mathrm{m}/\mathrm{s}$at an angle of $60.76°$above the +x-axis.

Step-by-step solution

Step 1. Understanding the concept of the horizontal range

The horizontal range is the distance covered in the horizontal direction before landing.

Step 2. Given data and assumptions

The horizontal range traveled upon reaching the landing point,$R=195\mathrm{m}$

The vertical displacement upon reaching the landing point,$H=135\mathrm{m}$

The time elapsed before the projectile reaches the landing point, t = 6.6 s

Let us consider the upward and rightward directions to be positive -Y and -X axes,

respectively and take the value of acceleration due to gravity as $g=-9.8\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Setting up equations for the horizontal motion

The initial velocity along the x-direction will be $v_0\cos \theta$.

Applying the second equation of motion for horizontal motion, you get:

$R=v_0\cos \theta t+0$(As acceleration along the x-direction is zero)

Substituting the values in the above equation, you get:

$195=v_0\cos \theta \times 6.6$

From the above equation, you get:

role="math" localid="1644991524048" $v_0\cos \theta =29.55\mathrm{m}/\mathrm{s}\dots \left(\mathrm{i}\right)$

Step 4. Setting up equations for the vertical motion

The initial velocity along the y-direction will be $v_{\circ }\sin \theta$.

Applying the second equation of motion for vertical motion, you get:

$H=v_0\sin \theta ·t+\frac{1}{2}\left(-g\right)t^2$

Substituting the values in the above equation, you get:

$135=v_0\sin \theta \times 6.6+\frac{1}{2}\left(-9.8\right)\times 6.6^2$

From the above equation, you get:

role="math" localid="1644991569420" $v_0\sin \theta =52.79\mathrm{m}/\mathrm{s}\dots \left(\mathrm{ii}\right)$

Step 5. Calculating the magnitude and direction of initial velocity

The magnitude of the initial velocity $v_0$can be written in terms of horizontal and vertical velocities as:

$v_0=\sqrt{{\left(v_0\sin \theta \right)}^2+{\left(v_0\cos \theta \right)}^2}$

Substituting the values from equations (i) and (ii), you get:

$\begin{array}{c}{v}_{\circ }=\sqrt{{52.79}^2+{29.55}^2}\\ =60.50\mathrm{m}/\mathrm{s}\end{array}$

The direction of initial velocity can be calculated by calculating the tangent function, as shown below.

$\begin{array}{c}\tan \theta =\frac{v_{\circ }\sin \theta }{v_{\circ }\cos \theta }\\ =\frac{52.79}{29.55}\end{array}$

Solving the above equation for $\theta$, you can write:

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{52.79}{29.55}\right)\\ =60.76°\end{array}$

Thus, the magnitude of the initial velocity is $60.50\mathrm{m}/\mathrm{s}$, and the direction of the initial velocity is $60.76°$above the +x-axis.