The angle of projection at which the horizontal range and maximum height of projectile are equal is

\\(R=\dfrac { { u }^{ 2 }\sin { 2\theta } }{ g } \\ H=\dfrac{ { u }^{ 2 }\sin { ^{ 2 }\theta } }{ 2g } \\ \dfrac{ { u }^{ 2 }\sin { 2\theta } }{ g } =\dfrac{ { u }^{ 2 }\sin { ^{ 2 }\theta } }{ 2g } \\ 2\sin { \theta } \cos { \theta } =\dfrac{ \sin { ^{ 2 }\theta } }{ 2 } \\ u\cos { \theta } =\sin { \theta } \\ \tan { \theta } =4\\ \theta =\tan ^{ -1 }{ 4 } \\)

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is :

Let time of flight be T then \\(T=\dfrac{u}{g}\\)Let h be the distance covered during last 't' second of its ascentVelocity at point \\(B=V_{B}=u-g(T-t)\\)\\(=u-g(\dfrac{u}{g}-t)=gt\\)\\(\Rightarrow h=V_{B}t-\dfrac{1}{2}gt^{2} \Rightarrow h=gt^{2}-\dfrac{1}{2}gt^{2}=\dfrac{1}{2}gt^{2}\\)

A projectile is given an initial velocity of \\((\hat i + 2\hat j)m/s\\) where \\(\hat i\\) is along the ground and \\(\hat j\\) is along the vertical. If \\(g = 10 m/s^2\\), the equation of its trajectory is

\\(U=(\hat{i}+2\hat{j}) m/0\\)\\(Ux=\hat{i} m/0\\)\\(x=S_x=Ux\cdot t\\)\\(x=S_x=1t=t\\) …………..(ii)\\(U_y=2\hat{j} m/0\\)\\(y=S_y-Uyt+\dfrac{1}{2}at^2\\)\\(S_y=2t-\dfrac{1}{2}\times 10t^2\\)\\(y=S_y=2t-5t^2\\) ………..(ii)From equation (i)\\(y=2(x)-5x^2\\)\\(5x^2-2x+y=0\\).

A cricketer can throw a ball to a maximum horizontal distance of \\(100\ m\\). How much high above the ground can the cricketer throw the same ball ?

Maximum horizontal distance, \\(R = 100 m\\) The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is \\(45^ \circ\\), i.e., \\(= 45 ^\circ\\).The max horizontal range for a projection velocity v is given by the relation:\\(R_{max}=\dfrac{u^2}{g}\\)\\(100= \dfrac{u^2}{g}\\) The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.Acceleration, \\(a = g\\)Using the third equation of motion:\\(v^2 - u^2= -2gH\\)\\(H = u^2 / 2g = 100 / 2 = 50 m\\)

A projectile is fired at angle of \\(45^{\circ}\\) with the horizontal. Elevation angle of the projectile at its heighest point as seen from the point of proejection, is

REF. Image.\\( \theta = 45^{\circ}\\) projection speed = uhorizontal velocity \\( ux = u cos 45^{\circ} = \frac{u}{\sqrt{2}}\\)initial vertical velocity uy \\(= u sin 40^{\circ} = \frac{u}{\sqrt{2}}\\)\\(height_{max}\\) attained H\\( = \frac{u^{2}sin^{2}\theta }{29} = \frac{u^{2}/2}{29}\\) \\( H = \frac{u^{2}}{49}\\)half of the range \\( = \frac{R}{2} = \frac{u^{2}sin2\theta }{29} = \frac{u^{2}sin 90^{\circ}}{29}\\)\\( \frac{R}{2} = \frac{u^{2}}{29}\\)elevation \\( \phi = tan^{-1} \frac{H}{R/2}\\)\\(\phi = tan^{-1}\frac{u^{2}/49}{u^{2}/29}\\)Or \\( \phi = tan^{-1} 0.5\\)

Show that the path of a projectile is a parabola

Let a body is projected with speed \\(u m/s\\) inclined \\(\theta \\) with horizontal line\\(.\\)Then\\(,\\) vertical component of \\(u,\\) \\({u_y} = u\cos \theta \\)Horizontal component of \\({u_x} = u\sin \theta \\)acceleration on horizontal\\(,\\) \\(ax=0\\)acceleration on vertical\\(,\\) \\(ay=-g\\)Now\\(,\\) use formula \\(X = {u_x}t\\)\\(X = u\cos \theta .t\\)\\(t = X/u\cos \theta \\)--------------------------\\((1)\\)Again\\(,\\) \\(y = {u_y}t + 1/2{a_y}{t^2}\\)\\(y = u\sin \theta t - 1/2g{t^2}\\)put equation \\((1)\\) here\\(,\\)\\(y = u\sin \theta \times x/u\cos \theta - 1/2g{t^2} \times {x^2}/{u^2}{\cos ^2}\theta \\)\\( = \tan \theta x - 1/2g{x^2}/{u^2}{\cos ^2}\\)\\(y = \tan \theta .x - 1/2g{x^2}/{u^2}{\cos ^2}\theta \\)this equation is similar to standard equation of parabola \\(y = a{x^2} + bx + c\\) her\\(,\\) \\(a,b\\)and \\(c\\) are constantSo\\(,\\) A projectile motion is parabolic motion\\(.\\)

A projectile can have the same range R for two angles pf projection. If \\(t_1\\) and \\(t_2\\) be the times of flight in the two cases, then the product of the two times of flight is directly proportional to

We know thatRange \\(R=\dfrac{4^2\sin 2\theta}{9}\\)If we put \\(\theta\rightarrow 90-\theta\\)We get\\(R=u^2\dfrac{\sin 2(90-\theta)}{9}=\dfrac{4^2\sin 180-2\theta}{29}\\)\\(R=\dfrac{4^2\sin 2\theta}{29}\\)Both ranges \\(\theta\\) & \\(90-\theta\\) are sameSo \\(\cos\theta\\) \\(T=\dfrac{24\sin\theta}{9}\\); time of f lightwe may write \\(t_1=\dfrac{24\sin\theta}{9}\\)\\(\theta \rightarrow 90-\theta\\)\\(t_2=\dfrac{24\sin(90-\theta)}{9}=\dfrac{24\cos\theta}{9}\\)\\(\Rightarrow t_1t_2=\dfrac{24^22\sin\theta \cos\theta}{9^2}\\)\\(=\dfrac{2}{9}\cdot \dfrac{4^2\sin2\theta}{9}=\dfrac{2}{9}R\\).\\(\Rightarrow t_1t_2 \propto R\\).

A particle of mass m is projected with velocity v making an angle of 45 with the horizontal from level ground. When the particle lands on the level ground the magnitude of the change in its momentum will be :

Change in momentum \\(=\Delta \vec{P}\\)\\(=\Delta P_{x}\hat{i}+\Delta P_{x}\hat{j}\\), since \\(\Delta P_{x}=0\\)\\(=\Delta P_{y}\hat{j}=\vec{P}_{fy}-\vec{P}_{iy}\\);\\(P_{iy}= mvsin 45^{},P_{fy} = mvsin 45^{}\\)\\(\Delta \vec{P}=-mvsin 45^{} - mvsin 45^{}\\)\\(=-2mvsin 45^{}=-\sqrt{2}mv_{j} = \left |\Delta \vec{P} \right |\\)\\(=\sqrt{2}mv\\)e sign shows the direction of change inmomentum in e y-direction.

A particle is projected with a velocity \\(v\\) such that its range on horizontal plane is twice the greatest height attained by it. The range of the projectile will be\\((g\\) is the acceleration due to gravity).

Given, Range, \\(R = 2H\\)But \\(R = 4H\cot \theta\\)\\(\Rightarrow \cot \theta = \dfrac {1}{2}\\)From triangle we can say that,\\(\sin \theta = \dfrac {2}{\sqrt {5}}\\) and \\(\cos \theta = \dfrac {1}{\sqrt {5}}\\)So, the range of the projectile\\(R = \dfrac {2v^{2} \sin \theta \cos \theta}{g} = \dfrac {2v^{2}}{g} \times \dfrac {2}{\sqrt {5}}\times \dfrac {1}{\sqrt {5}}\\)\\(= \dfrac {4v^{2}}{5g}\\).

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is

Let \\(v\\) be velocity of a projectile at maximum height \\( H\\)v = ucos\\(\theta\\)According to given problem, v = \\(\displaystyle \dfrac{u}{2}\\)\\(\displaystyle \therefore \dfrac{u}{2} = ucos \theta \Rightarrow cos\theta = \dfrac{1}{2}\\)\\(\displaystyle \theta = 60^o\\)

Maximum Height, Time of Flight and Horizontal Range of a Projectile

definition

Average Velocity of Projectile

The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with

u

making an angle

θ

with the vertical.
There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum

= 0

It is the horizontal velocity which is uniform and hence

v_{a v} = u_{x} = u cos θ

For a general point:
Displacement in Y-direction:

y = u s i n θ \times t - \frac{g t ^{2}}{2}

Displacement in X-direction:

x = u c o s θ \times t

Now in order to calculate average velocity:
Average Velocity =

\frac{N e t D i s p l a c e m e n t}{T o t a l t i m e}

formula

Range of a Projectile Motion

The horizontal distance travelled by a projectile from its initial position

(x = y = 0)

to the position where it passes

y = 0

during its fall is called the horizontal range,

R

. It is the distance travelled during the time of flight

T_{f}

. therefore, the range

R

R = (v_{o} cos θ_{o}) (T_{f}) = (v_{o} cos θ_{o}) (2 v_{o} sin θ_{o}) / g

......(1)
Or,

R = \frac{v _{o} ^{2} sin 2 θ _{o}}{g}

......(2)
Equation 2 shows that for a given projectile velocity

v_{o}

R

is maximum when

sin 2 θ_{o}

is maximum, i.e. when

θ_{o} = 4 5^{o}

.
The maximum horizontal range is, therefore

R_{m} = \frac{v _{0} ^{2}}{g}

diagram

Time of Flight of a Projectile

Let, time taken to reach maximum height

= t_{m}

Now,

v_{x} = v_{o} cos θ_{o}

and

v_{y} = v_{o} sin θ_{o} - g t

Since, at this point,

v_{y} = 0

, we have:

v_{o} sin θ_{o} - g t_{m} = 0

Or,

t_{m} = (v_{o} sin θ_{o}) / g

Therefore, time of flight

= T_{f} = 2 t_{m} - 2 (v_{o} sin θ_{o}) / g

because of symmetry of the parabolic path.

diagram

Maximum Height of a Projectile

x = v_{x} t = (v_{o} cos θ_{o}) t

and

y = (v_{o} sin θ_{o}) t - (1 / 2) g t^{2}

The maximum height

h_{m}

is given by:

y = h_{m} = (v_{o} sin θ_{o}) (\frac{v _{o} sin θ _{o}}{g}) - \frac{g}{2} (\frac{v _{o} sin θ _{o}}{g})^{2}

(for

t = t_{m}

)
Or,

h_{m} = \frac{( v _{o} sin θ _{o} ) ^{2}}{2 g}

Maximum Height, Time of Flight and Horizontal Range of a Projectile

definition

Average Velocity of Projectile

formula

Range of a Projectile Motion

diagram

Time of Flight of a Projectile

diagram

Maximum Height of a Projectile

REVISE WITH CONCEPTS

Equations of Motion for a Projectile

Velocity and Direction of Motion at Different Location of Projectile Motion

Trajectory of a Projectile

Equations of Motion for a Projectile - Problem L2

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