Question

plzz Answer this plz don't spam ​

Answer 1

Answer:

• The initial velocity (u) is 12.5 m/s

Given:

1. Trajectory => y = 0.5 x - 0.04 x²

Explanation:

$\rule{300}{1.5}$

The given equation is, y = 0.5 x - 0.04 x².

Comparing it with the trajectory equation.

$\longrightarrow\sf y=(\tan \theta)\;x- \Bigg\langle\dfrac{g}{2\;u^{2}\cos^{2}\theta}\Bigg\rangle\;x^{2}$

So, from comparing it we get,

• tanθ = 0.5
• g/(2 u² cos²θ) = 0.04

Now, taking first case

$\longrightarrow\sf \tan \theta=0.5\\\\\\\longrightarrow\sf \tan \theta =\dfrac{5}{10}\\\\\\\longrightarrow\sf \tan \theta =\dfrac{1}{2}\\\\\\\longrightarrow\sf \tan \theta =\dfrac{1}{2}=\dfrac{Opp.\;side}{Adjacent\;side}=\dfrac{BC}{AB}$

Now, making Right angle triangle.

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(0,0){\line(1,0){4}}\put(0,0){\line(1,1){4}}\qbezier(0.71,0)(0.95,0.35)(0.35,0.35)\put(4,0){\line(0,1){4}}\put(1,0.4){ \theta }\put(4.3,2){ \sf 1 }\put(2,-0.5){ \sf 2 }\put(0,-0.5){ \sf A }\put(4,-0.5){ \sf B }\put(4,4){ \sf C }\end{picture}$

From Pythagoras theorem.

⇒ AC² = AB² + BC²

Substituting the values,

⇒ AC² = (2)² + (1)²

⇒ AC² = 4 + 1

⇒ AC² = 5

⇒ AC = √5

∴ We got AC (Hypotenuse) value.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, second case.

$\longrightarrow\sf \dfrac{g}{2\;u^{2}\cos^{2}\theta}=0.04\\\\\\\longrightarrow\sf 2\;u^{2}\cos^{2}\theta=\dfrac{g}{0.04}\\\\\\\longrightarrow\sf 2\;(u\cos \theta)^{2}=\dfrac{10\times 100}{4}}\\\\\\\longrightarrow\sf 2\;(u\cos \theta)^{2}=250\\\\\\\longrightarrow\sf (u\cos \theta)^{2}=\dfrac{250}{2}\\\\\\\longrightarrow\sf (u\cos \theta)^{2}=125$

$\bullet\qquad\boxed{\sf\cos \theta = \dfrac{AB}{AC}=\dfrac{2}{\sqrt{5}}}$

$\longrightarrow\sf \Bigg\langle u\times \dfrac{2}{\sqrt{5}}\Bigg\rangle^{2}=125\\\\\\\longrightarrow\sf u^{2}\times \dfrac{4}{5}=125\\\\\\\longrightarrow\sf u^{2}=\dfrac{125\times 5}{4}\\\\\\\longrightarrow\sf u^{2}=\dfrac{625}{4}\\\\\\\longrightarrow\sf u=\sqrt{\dfrac{625}{4}}\\\\\\\longrightarrow\sf u=\dfrac{25}{2}$

$\longrightarrow\sf u=12.5\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf u=12.5\;m/s}}}}$

∴ The initial velocity (u) is 12.5 m/s.

Hence Option-3 is correct!

$\rule{300}{1.5}$

Answer 2

Explanation:

option 3 is correct.

hope it helps you