Trajectory of particle in a projectile motion is given as y = x - x280

1.	Trajectory of particle in a projectile motion is given as y = x - x280 . Here , x and y are in meters. For this projectile motion match the following with g = 10 m / s2 "coloumn1coloumn2(a)Angle of Projection(p)20(b)Angle of velocity with horizontal after 4s(q)80(c)Maximum Hieght(r)45∘(d)Horizontal Range(s)tan−112 "
Answer» 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 / $s^{2}$ " $\begin{matrix} c o l o u m n 1 & c o l o u m n 2 (a) A n g l e o f P r o j e c t i o n & (p) 20 (b) A n g l e o f v e l o c i t y w i t h h o r i z o n t a l a f t e r 4 s & (q) 80 (c) M a x i m u m H i e g h t & (r) 45^{\circ} (d) H o r i z o n t a l R a n g e & (s) {t a n}^{- 1} \frac{1}{2} \end{matrix}$ "

Trajectory of particle in a projectile motion is given as y = x - x280 . Here , x and y are in meters. For this projectile motion match the following with g = 10 m / s2 "coloumn1coloumn2(a)Angle of Projection(p)20(b)Angle of velocity with horizontal after 4s(q)80(c)Maximum Hieght(r)45∘(d)Horizontal Range(s)tan−112 "

Answer»

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 / $s^{2}$

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

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