Question

A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35.0 with the horizontal, as shown inthe figure. 

(a) Determine the time taken by the
projectile to hit point P at ground level. (b)
Determine the distance X of point P from the base of
the vertical cliff. At the instant just before the
projectile hits point P, find (c) the horizontal and the
vertical components of its velocity, (d) the magnitude
of the velocity, and (e) the angle made by the
velocity vector with the horizontal. (f) Find the
maximum height above the cliff top reached by the
projectile

Answer 1

Concepts and reason

The projectile motion can be defined as the motion of a particle when thrown, follows the curved path because of gravity.

Time taken by the projectile motion can be found out using the kinematics equations.

The horizontal range can be found out using the kinematics equations.

The magnitude of the velocity can be found out using the x-component and the y-component of the velocity.

The angle made by the velocity vector can be found out using the x-component and the y-component of the velocity.

Fundamentals

The x-component and the y-component of the initial velocity is,

$\begin{array}{c}\\{v_{0x}} = {v_0}\cos \theta \\\\{v_{0y}} = {v_0}\sin \theta \\\end{array}$

Here, ${v_0}$ is the initial velocity, ${v_{0x}}$is the x-component of the velocity, ${v_{0y}}$ is the y-component of the velocity, and $\theta$ is the angle between both the components of the velocity.

The vertical distance travelled by the projectile to hit the point on the ground is,

$y = {y_0} + {v_{0y}}t + \frac{1}{2}g{t^2}$

Here, y is the distance travelled to hit the point on the ground, ${y_0}$ is the initial distance, ${v_{0y}}$ is the y-component of velocity, t is the time taken to hit the ground, and g is the acceleration due to gravity.

The horizontal range of the object when it hit the point on the ground is,

$x = {x_0} + {v_{0x}}t$

Here, $x$ is the horizontal distance, ${x_0}$is the initial distance, and ${v_{0x}}$is the horizontal component of velocity, and t is the time.

The horizontal component of velocity is,

${v_x} = {v_{0x}}$

Here, ${v_x}$ is the horizontal component of velocity.

The vertical component of velocity is,

${v_y} = {v_{0y}} + gt$

Here, ${v_y}$ is the vertical component of velocity, g is the acceleration due to gravity, and t is the time.

The magnitude of the velocity is,

$\left| v \right| = \sqrt {{v_x}^2 + {v_y}^2}$

Here, $\left| v \right|$ is the magnitude of the velocity.

The angle made by the velocity vector is,

$\tan \theta = \frac{{{v_y}}}{{{v_x}}}$

Here, $\theta$ is the angle between the x-component and y-component of the velocity.

The maximum height reached by the projectile is,

${h_{{\rm{max}}}} = \frac{{{{\left( {{v_0}\sin \theta } \right)}^2}}}{{2g}}$

Here, ${h_{{\rm{max}}}}$ is the maximum height reached by the projectile, ${v_0}$ is the initial velocity, and g is the acceleration due to gravity.

(a)

Consider the formula for the vertical distance travelled by the object to hit the ground at the point p.

The distance travelled by the projectile to hit the point on the ground is,

$y = {y_0} + {v_{0y}}t + \frac{1}{2}g{t^2}$

Substitute ${v_0}\sin \theta$ for${v_{0y}}$.

$y = {y_0} + {v_0}\sin \theta t + \frac{1}{2}g{t^2}$

Substitute 0 for y, 115 m for${y_0}$, 65 m/s for ${v_0}$, $35^\circ$ for$\theta$, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g.

$\begin{array}{c}\\0 = \left( {115{\rm{ m}}} \right) + \left( {65{\rm{ m/s}}} \right)\sin \left( {35^\circ } \right)t + \frac{1}{2}\left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right){t^2}\\\\0 = 115 + 37t - 4.9{t^2}\\\end{array}$

The above equation reduces to quadratic form.

$4.9{t^2} - 37t - 115 = 0$

Compare the quadratic equation with the following quadratic equation and find the value of time.

$a{t^2} + bt + c = 0$

Solve the quadratic equation.

$t = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substitute $4.9{\rm{ m/}}{{\rm{s}}^2}$ for a, $- 37{\rm{ m/s}}$ for b, and -115 m for c.

$\begin{array}{c}\\t = \frac{{ - \left( { - 37{\rm{ m/s}}} \right) \pm \sqrt {{{\left( { - 37{\rm{ m/s}}} \right)}^2} - 4\left( {4.9{\rm{ m/}}{{\rm{s}}^2}} \right)\left( { - 115{\rm{ m}}} \right)} }}{{2\left( {4.9{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\t' = - 2.36{\rm{ s}}\\\\t{\rm{ = 9}}{\rm{.92 s}}\\\end{array}$

The quadratic equation of t will give two values for t. Consider only the positive value of the time because negative value is not possible.

Thus, the time taken by the projectile to hit point P on the ground is 9.92 s.

(b)

Determine the horizontal range for the horizontal motion.

The horizontal range of the object when it hit the point on the ground is,

$x = {x_0} + {v_{0x}}t$

Substitute ${v_0}\cos \theta$ for${v_{0x}}$.

$x = {x_0} + {v_0}\cos \theta t$

Substitute 0 for ${x_0}$, 65 m/s for ${v_0}$, $35^\circ$ for $\theta$, and 9.92 s for t in $x = {x_0} + {v_0}\cos \theta t$.

$\begin{array}{c}\\x = 0{\rm{ m}} + \left( {65{\rm{ m/s}}} \right)\cos \left( {35^\circ } \right)\left( {9.92{\rm{ s}}} \right)\\\\ = 528{\rm{ m}}\\\end{array}$

The horizontal range is 528 m.

(c)

The x-component and the y-component of the velocity is,

${v_x} = {v_{0x}}$

Substitute ${v_0}\cos \theta$ for${v_{0x}}$.

${v_x} = {v_0}\cos \theta$

Substitute 65 m/s for ${v_0}$ and $35^\circ$ for$\theta$ in above equation as follows:

$\begin{array}{c}\\{v_x} = \left( {65{\rm{ m/s}}} \right)\cos 35^\circ \\\\ = 53.24{\rm{ m/s}}\\\end{array}$

The vertical component of velocity is,

${v_y} = {v_{0y}} + gt$

Here, ${v_y}$ is the vertical component of velocity, g is the acceleration due to gravity, and t is the time.

Substitute ${v_0}\sin \theta$ for${v_{0y}}$.

${v_y} = {v_0}\sin \theta + gt$

Substitute 65 m/s for${v_0}$, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 9.92 s for t, and $35^\circ$ for $\theta$in ${v_y} = {v_0}\sin \theta + gt$.

$\begin{array}{c}\\{v_y} = \left( {65{\rm{ m/s}}} \right)\sin 35^\circ - \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {9.92{\rm{ s}}} \right)\\\\ = - 59.9{\rm{ m/s}}\\\end{array}$

The horizontal component of velocity and the vertical component of velocity is 53.24 m/s and -59.9 m/s respectively.

(d)

Consider the expression of the magnitude of the velocity vector.

The magnitude of the velocity is,

$\left| v \right| = \sqrt {{v_x}^2 + {v_y}^2}$

Substitute 53.24 m/s for ${v_x}$ and -59.9 m/s for ${v_y}$.

$\begin{array}{c}\\\left| v \right| = \sqrt {{{\left( {53.24{\rm{ m/s}}} \right)}^2} + {{\left( { - 59.9{\rm{ m/s}}} \right)}^2}} \\\\ = 80.1{\rm{ m/s}}\\\end{array}$

The magnitude of velocity is 80.1 m/s.

(e)

Consider the expression of angle made by the velocity vector.

The angle made by the velocity vector is,

$\tan \theta = \frac{{{v_y}}}{{{v_x}}}$

Substitute 53.24 m/s for ${v_x}$ and -59.9 m/s for${v_y}$.

$\begin{array}{c}\\\tan \theta = - \frac{{59.9{\rm{ m/s}}}}{{53.24{\rm{ m/s}}}}\\\\\theta = {\tan ^{ - 1}}\left( { - \frac{{59.9{\rm{ m/s}}}}{{53.24{\rm{ m/s}}}}} \right)\\\\ = - 48.4^\circ \\\end{array}$

The angle made by the velocity vector is$- 48.4^\circ$.

(f)

Consider the expression of the maximum height of the projectile.

The maximum height reached by the projectile is,

${h_{{\rm{max}}}} = \frac{{{{\left( {{v_0}\sin \theta } \right)}^2}}}{{2g}}$

Substitute 65 m/s for ${v_0}$, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $35^\circ$ for $\theta$.

$\begin{array}{c}\\{h_{{\rm{max}}}} = \frac{{{{\left( {\left( {65{\rm{ m/s}}} \right)\sin \left( {35^\circ } \right)} \right)}^2}}}{{2\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 70.9{\rm{ m}}\\\end{array}$

The maximum height is 70.9 m.

Ans: Part a

The time taken by the projectile to hit point P on the ground is 9.92 s.

Part b

The horizontal range is 528 m.

Part c

The horizontal component of velocity is 53.24 m/s and the vertical component of velocity is -59.9 m/s.

Part d

The magnitude of velocity is 80.1 m/s.

Part e

The angle made by the velocity vector is$- 48.4^\circ$.

Part f

The maximum height above the cliff top reached by the projectile is 70.9 m.