1 answer

Projectile

Question:

In Fig. 4-33, a stone is projected at a cliff of height hwith an initial speed of 42.0 m/sdirected 60.0° above the horizontal. The stone strikes atA,5.50 s after launching.

Figure 4-33

(a) Find the height, h, of the cliff.
m
(b) Find the speed of the stone just before impact atA.
m/s
(c) Find the maximum height H reached above theground.
m

Answers

Concepts and reason

The concepts used to solve the question is the projectile motion and the equations of the motion.

At first, determine the horizontal and the vertical components of the velocity and then apply the equations of the motion to determine the height of the cliff, speed of the stone and maximum height reached by the stone from the ground.

Fundamentals

Projectile motion

A projectile is any object upon which only force acts the force of gravity. When a stone is thrown upwards, the force of the gravity is acting downwards. Thus, the object is moving upwards and slowing down.

Hence, the force of gravity causes a vertical acceleration which pulls the object in a downward direction.

Equations of motion

The equations describing the motion of the objects with respect to the time are defined as the kinematic equations. These equations are used when there is constant acceleration.

Constant acceleration means that the speed of the object is changing uniformly. Constant acceleration does not mean that the speed of the object is constant.

If the speed of the object is constant, then there will be zero acceleration.

The equations of the motion are described below.

v=u+atv = u + at

Here, vv is the final velocity, uu is the initial velocity, aa is the acceleration of the object and tt is the time taken.

The displacement equation is,

S=ut+at22S = ut + \frac{{a{t^2}}}{2}

And

v2=u2+2aS{v^2} = {u^2} + 2aS

(a)

The vertical component of the velocity of stone is,

vy=v0sinθ{v_{\rm{y}}} = {v_0}\sin \theta …… (1)

Here, v0{v_0} is the initial velocity of the stone and θ\theta is the angle of the projection.

Substitute42m/s42{\rm{ m/s}} for v0{v_0} and 6060^\circ for θ\theta in the equation (1).

vy=v0sinθ=(42m/s)sin60=36.373m/s\begin{array}{c}\\{v_{\rm{y}}} = {v_0}\sin \theta \\\\ = \left( {42{\rm{ m/s}}} \right)\sin 60^\circ \\\\ = 36.373{\rm{ m/s}}\\\end{array}

The horizontal component of the velocity of the stone is,

vx=v0cosθ{v_{\rm{x}}} = {v_0}\cos \theta …… (2)

Substitute42m/s42{\rm{ m/s}} for v0{v_0} and 6060^\circ for θ\theta in the equation (2).

vx=v0cosθ=(42m/s)cos60=21m/s\begin{array}{c}\\{v_{\rm{x}}} = {v_0}\cos \theta \\\\ = \left( {42{\rm{ m/s}}} \right)\cos 60^\circ \\\\ = 21{\rm{ m/s}}\\\end{array}

Apply the equation of the motion to determine the height of the cliff.

h=vytgt22h = {v_{\rm{y}}}t - \frac{{g{t^2}}}{2} …… (3)

Here, gg is the acceleration due to gravity.

Substitute, 36.373m/s36.373{\rm{ m/s}} forvy{v_{\rm{y}}}, 5.50s5.50{\rm{ s}} for tt and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg in the equation (3).

h=(36.373m/s)(5.50s)(9.81m/s2)(5.50s)22=51.675m\begin{array}{c}\\h = \left( {36.373{\rm{ m/s}}} \right)\left( {5.50{\rm{ s}}} \right) - \frac{{\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right){{\left( {5.50{\rm{ s}}} \right)}^2}}}{2}\\\\ = 51.675{\rm{ m}}\\\end{array}

(b)

The final velocity along the horizontal direction is,

vfx=vx=21m/s\begin{array}{c}\\{v_{{\rm{f}}x}} = {v_{\rm{x}}}\\\\ = 21{\rm{ m/s}}\\\end{array}

The final velocity of the stone along the vertical direction is,

vfy=vygt{v_{{\rm{fy}}}} = {v_{\rm{y}}} - gt …… (4)

Substitute, 36.373m/s36.373{\rm{ m/s}}for vy{v_{\rm{y}}}5.50s5.50{\rm{ s}}tt and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg in the equation (4).

vfy=vygt=(36.373m/s)(9.81m/s2)(5.50s)=17.582m/s\begin{array}{c}\\{v_{{\rm{fy}}}} = {v_{\rm{y}}} - gt\\\\ = \left( {36.373{\rm{ m/s}}} \right) - \left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {5.50{\rm{ s}}} \right)\\\\ = - 17.582{\rm{ m/s}}\\\end{array}

The speed of the stone just before the impact at point A is,

v=vfx2+vfy2=(21m/s)2+(17.582m/s)2=27.388m/s\begin{array}{c}\\v = \sqrt {{v_{{\rm{fx}}}}^2 + {v_{{\rm{fy}}}}^2} \\\\ = \sqrt {{{\left( {21{\rm{ m/s}}} \right)}^2} + {{\left( { - 17.582{\rm{ m/s}}} \right)}^2}} \\\\ = 27.388{\rm{ m/s}}\\\end{array}

(c)

The maximum height of the stone is,

H=vfy2vy22gH = \frac{{{v_{{\rm{fy}}}}^2 - {v_{\rm{y}}}^2}}{{ - 2g}} …… (5)

Substitute0m/s0{\rm{ m/s}}forvfy{v_{{\rm{fy}}}}, 36.373m/s36.373{\rm{ m/s}} for vy{v_{\rm{y}}} and 9.81m/s29.81{\rm{ m/}}{{\rm{s}}^2} for gg in the equation (5).

H=vfy2vy22g=(0m/s)2(36.373m/s)22(9.81m/s2)=67.430m\begin{array}{c}\\H = \frac{{{v_{{\rm{fy}}}}^2 - {v_{\rm{y}}}^2}}{{ - 2g}}\\\\ = \frac{{{{\left( {0{\rm{ m/s}}} \right)}^2} - {{\left( {36.373{\rm{ m/s}}} \right)}^2}}}{{ - 2\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 67.430{\rm{ m}}\\\end{array}

Ans: Part a

The height of the cliff is 51.675m{\bf{51}}{\bf{.675 m}}

Part b

The speed of the stone just before the impact is 27.388m/s{\bf{27}}{\bf{.388 m/s}}

Part c

The maximum height reached by the stone above the ground is 67.430m{\bf{67}}{\bf{.430 m}}

.

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