Trajectory of particle in a projectile motion is given as y - x - x 2-80- Here- x and y are in metre- For this projectile motion match the following and select proper option g -10 m - s 2 - beginarray - Wline- text with horizontal -text C - text Maximum height -text r - 45- lcirc llthlineendarray A- A - r - B - r - C - q - D - pB- A - s - B - r - C - p - D - qC- A - s - B - s - C - q - D - pD- A - r - B - r - C - p - D - q

Given y=x ( 1 x80 ) Comparing with the standard form y=x tan θ ( 1 xR ), ⇒tan θ = 1 ⇒θ = 45^∘ and R= 80 m We have, 4H = R tan θ 4H =80(1) ⇒ H = 20 m ⇒u_y^22 ...

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Standard XII

Physics

The Equation for the Path of Projectile

Trajectory of...

Question

Trajectory of particle in a projectile motion is given as $y = x - \frac{x^{2}}{80}$ . Here, $x$ and $y$ are in $metre$ . For this projectile motion match the following and select proper option $(g = 10 {m/s}^{2})$ :
$\begin{matrix} Column I & Column II (A) & Angle of projection & (p) & 20 m (B) & Angle of velocity & (q) & 80 m with horizontal after 4 s (C) & Maximum height & (r) & 45^{\circ} (D) & Horizontal range & (s) & {tan}^{- 1} (\frac{1}{2}) \end{matrix}$

A \to s, B \to s, C \to q, D \to p

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A \to r, B \to r, C \to q, D \to p

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A \to r, B \to r, C \to p, D \to q

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A \to s, B \to r, C \to p, D \to q

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Solution

The correct option is C $A \to r, B \to r, C \to p, D \to q$
Given
$y = x (1 - \frac{x}{80})$

Comparing with the standard form $y = x tan θ (1 - \frac{x}{R})$ ,

$\Rightarrow tan θ = 1$
$\Rightarrow θ = 45^{\circ}$
and $R = 80 m$

We have,
$4 H = R tan θ$
$4 H = 80 (1)$
$\Rightarrow H = 20 m$

$\Rightarrow \frac{u_{y}^{2}}{2 g} = 20$
$\Rightarrow u_{y} = 20$

Since, $tan θ = 1$
$\Rightarrow u_{x} = u_{y} = 20$

At $t seconds$ ,
$tan α = \frac{u_{y} - g t}{u_{x}}$
at $t = 4 s$
$\Rightarrow tan α = \frac{20 - 40}{20}$
$\Rightarrow α = {tan}^{- 1} (- 1) = 45^{\circ}$ below the horizontal.

Hence, option $(C)$ is the right choice.

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Similar questions

Trajectory of particle in a projectile motion is given as y = x - $\frac{x^{2}}{80}$ .

Here , x and y are in meters. For this projectile motion match the following with g = 10 m / $s^{2}$

" $\begin{matrix} c o l o u m n 1 & c o l o u m n 2 (a) A n g l e o f P r o j e c t i o n & (p) 20 (b) A n g l e o f v e l o c i t y w i t h h o r i z o n t a l a f t e r 4 s & (q) 80 (c) M a x i m u m H i e g h t & (r) 45^{\circ} (d) H o r i z o n t a l R a n g e & (s) {t a n}^{- 1} \frac{1}{2} \end{matrix}$ "