Question

do it with explanation ​

Answer 1

✫ Projectile Motion ✫

• An object thrown at an angle is called a projectile, and its motion is called projectile motion.

• A Projectile traces out a parabolic path, owing to the constant gravitational force acting downwards.

• Suppose we have a projectile which we throw with an initial velocity u  and projection angle θ.

✫ Required Diagram ✫

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}$

• We can resolve the Projectile Motion into 2 Dimensions. Let's consider the two Cartesian Coordinate Axes -  the X and Y axes. Consider the Origin to be at the point of projection.

It is easy to see that gravitational acceleration g is in the negative Y direction.

We can resolve the initial velocity u in two components:

$\sf u_x = u\, cos\, \theta \\\\\sf u_y = u\, sin\,\theta$

Let's assume x and y to be the displacements in X and Y directions respectively.

There's no acceleration in the X direction. So can write

$\sf Distance = Speed \times Time \\\\\sf \implies x = u_x\times t \\\\\sf \implies x = u\, cos\, \theta \ t$

In the Y-direction, there's a uniform acceleration g along the negative Y-axis. So from the equations of uniformly accelerated motion, we have:

$\displaystyle\sf s = ut+\frac{1}{2}at^2 \\\\\\\sf \implies y = u_yt+\frac{1}{2}(-g)t^2 \\\\\\\sf \implies y=u\, sin\, \theta \ t -\frac{1}{2}gt^2$

Let's call Total Time of Flight as T.

The flight of projectile ends when it touches the ground (i.e. the y-coordinate becomes 0)

So, by putting y=0 we can get the Time of Flight.

Put y=0:

$\sf\displaystyle 0=u\, sin\, \theta \ t -\frac{1}{2}gt^2 \\\\\\\sf\implies gt^2-2u\, sin \, \theta\ t = 0 \\\\\\\sf \implies t(gt-2u\, sin\, \theta)=0 \\\\\\\sf \implies t=0\quad OR \quad t = \frac{2u\, sin\,\theta}{g}$

t=0 corresponds to the time of projection. The other solution corresponds to the time when the projectile hits the ground again. That gives us the Total Time of Flight.

$\sf\displaystyle \implies T = \frac{2u\, sin\, \theta}{g}$

The projectile first goes up, reaches a maximum height and then comes back down. It follows a symmetrical path.

The time taken for it reach Maximum Height is the same as it takes to reach back to the ground from Maximum Height.

So, at time $\sf t=\frac{T}{2}$, the body reaches the Maximum Height. Let's call Maximum Height as H.

$\sf\displaystyle \rightarrow y=u\, sin\, \theta \ t -\frac{1}{2}gt^2 \\\\\\\sf \textsf{y=H when } t=\frac{T}{2}=\frac{u\, sin\, \theta}{g} \\\\\\ \sf\implies H = u\, sin\, \theta \left(\frac{u\, sin\, \theta}{g}\right)-\frac{1}{2}g\left(\frac{u\, sin\, \theta}{g}\right)^2 \\\\\\\sf \implies H=\frac{u^2\, sin^2\,\theta}{g}-\frac{u^2\, sin^2\,\theta}{2g} \\\\\\ \sf\implies \boxed{\sf H=\frac{u^2\, sin^2\,\theta}{2g}}$

Maximum Horizontal Length covered is called Range (R). We can find Range R by getting the value of x-coordinate when the projectile hits the ground again.

$\sf\displaystyle \rightarrow x=u\, cos\, \theta\ t \\\\\\\sf x=R \textsf{ when } t=T=\frac{2u\, sin\, \theta}{g} \\\\\\\sf \implies R = u\, cos\, \theta\ \left(\frac{2u\, sin\, \theta}{g}\right) \\\\\\\sf \implies \boxed{\sf R=\frac{2u^2\, sin\theta\, cos\, \theta}{g}}$

$\rule{300}{1}$

Now, We are given that Maximum Height = Range in the question. So, we put H=R.

$\sf\displaystyle R=H \\\\\\\sf \implies \frac{2\cancel{u^2}\, \cancel{sin\,\theta}\,cos\,\theta}{\cancel{g}}=\frac{\cancel{u^2}\,sin^{\cancel{2}}\,\theta}{2\cancel{g}} \\\\\\ \sf\implies 2\, cos\,\theta=\frac{sin\,\theta}{2} \\\\\\\sf \implies \frac{sin\,\theta}{cos\,\theta}=2\times 2 \\\\\\\sf \implies \boxed{\sf tan\,\theta=4}$

We want to find the Maximum Height, and we see that the formula for H contains $\sf sin^2\,\theta$. We must find this value.

$\sf\displaystyle tan\,\theta=4 \\\\\\\sf \rightarrow sec^2\,\theta=1+tan^2\,\theta \\\\\\\sf \implies sec^2\,\theta=1+4^2\\\\\\\sf \implies sec^2\,\theta=17 \\\\\\\sf \implies cos^2\,\theta=\frac{1}{sec^2\,\theta}=\frac{1}{17} \\\\\\\\ \sf\rightarrow sin^2\,\theta=1-cos^2\,\theta\\\\\\ \sf\implies sin^2\,\theta=1-\frac{1}{17} \\\\\\ \implies \boxed{\sf sin^2\,\theta=\frac{16}{17}}$

Now, we can easily get the expression for Maximum Height.

$\sf\displaystyle \rightarrow H=\frac{u^2\, sin^2\,\theta}{2g} \\\\\\\sf \implies H=\frac{u^2}{2g} \times \frac{16}{17} \\\\\\\\ \implies \huge \boxed{\boxed{\sf H=\frac{8u^2}{17g}}}$

$\rule{300}{1}$

Thus, If a body is projected with a velocity u such that its maximum height and range are same, then the maximum height is $\bold{\frac{8u^2}{17g}}$, where g is gravitational acceleration.

Answer 2

Explanation:

Lets consider u= initial speed of the projectile

θ= angle of projection

According to question,

H=R

2g

u

2

sin

2

θ

=

g

u

2

sin2θ

sinθ.sinθ=2×2sinθ.cosθ

sinθ=4cosθ

tanθ=4

As we know that, 1+tan

2

θ=sec

2

θ

cos

2

θ

1

=1+16=17

cos

2

θ=

17

1

sin

2

θ+cos

2

θ=1

sin

2

θ=1−cos

2

θ

sin

2

θ=1−

17

1

=

17

16

Maximum height, H=

2g

u

2

sin

2

θ

H=

2g

u

2

17

16

H=

17g

8u

2