Trajectory of particle in a projectile motion is given as y - x - x 2-80- Here- x and y are in meters- For this projectile motion match the following with g -10 m - s 2 -beginarray - c - c - Whine coloumn1 - coloumn2lllhlineaAngle ispace of Projection - p 20 lllhlinecMaximum ispace Hieght - r 45- lcirc lWhlinelendarrayA- a - r b - s c - p d - q B- a -pb -qc -rd -sC- a - r b - r c - p d - q D- a -pb -pc -rd -r

The correct option is C Comparing with the standard equation of projectile, y = x tan θ nbsp; gx^2/2 u^2cos^2 nbsp;θ We get , θ = nbsp; 45^∘ and u = 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Introduction to Projectile Motion

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 meters. For this projectile motion match the following with g = 10 m / $s^{2}$

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

Comparing with the standard equation of projectile,

y = x tan $θ$ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

We get , $θ$ = $45^{\circ}$ and u = 20 $\sqrt{2}$ m/s

Time period of this projectile is 4s . Hence after 4s velocity vector will again make $45^{\circ}$ with horizontal

Suggest Corrections

Similar questions

Q. 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}

Q. 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}

Let B and C be points (12, 19) and (23, 20) respectively. Let A $(α, β)$ be a point such that area of triangle formed by the points A, B and C is 70. Also the slope of the median through A is – 5.

Match the following:

\begin{matrix} C o l u m n I & C o l u m n I I (a) & A p o s s i b l e v a l u e α + β i s & (p) & 2 (b) & A p o s s i b l e v a l u e β - 2 α i s & (q) & 27 (c) & A p o s s i b l e v a l u e 3 α - β - 11 i s & (r) & - 33 (d) & A p o s s i b l e v a l u e α - 2 β + 16 i s & (s) & 47 (t) & 28 \end{matrix}

Three rods of material x and two rods of material y are connected as shown in figure. All the rods are of identical length and cross - sectional area. The end A is maintained at 100 $^{\circ}$ C and the junction C at 0 $^{\circ}$ C. It is given that resistance of rod of material x is R. Further, $K_{x}$ = ${2 K}_{y}$ .

Match the entries of column I and column II

$\begin{matrix} c o l u m n 1 & c o l u m n 2 A . T e m p e r a t u r e o f j u n c t i o n B & P . D o e s n o t b e h a v e l i k e a b a l a n c e d w h e a t s t o n e s^{'} s b r i d g e B . T e m p e r a t u r e o f j u n c t i o n C & Q . I s t w i c e o f t h e o t h e r C . T h i s n e t w o r k i s & R . {\frac{400}{7}}^{\circ} C D . R e s i s t a n c e o f r o d o f m a t e r i a l y & S . {\frac{300}{7}}^{\circ} C \end{matrix}$

Q. A body of mass

m_{1}

elastically collide with another stationary body of mass

m_{2}

\begin{matrix} Column - I & Column -II (a) After the collision, velocity of second & (p) > 1 body is maximum when \frac{m_{1}}{m_{2}} is (b) After the collision, the momentum & (q) < 1 of second body is maximum when \frac{m_{1}}{m_{2}} is (c) After the collision, K.E. of second & (r) = 1 body is maximum when \frac{m_{1}}{m_{2}} is (D) For elastic collision, coefficient of & (s) = 0 restitution is \end{matrix}

Join BYJU'S Learning Program

The correct option is D Comparing with the standard equation of projectile, y = x tan θ - gx22u2cos2θ We get , θ = 45∘ and u = 20 √2 m/s Time period of this projectile is 4s . Hence after 4s velocity vector will again make 45∘ with horizontal