Question

A body is allowed to slide from the top along a smooth inclined plane of length
5m at an angle of inclination 30°. If g=10ms?, time taken by the body to reach
the bottom of the plane is​

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf Time(t)=1.41\:sec}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Lenght (s) = 5m

• Angle of inclination $\sf{(\theta)}$ = 30°

• g = $\sf{10m/s^2}$

• Initially at rest (u) = 0

$\large\underline\pink{\sf To\:Find: }$

• Time taken to reach the bottom of the plane is (t) = ?

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Acceleration (a)

$\large{\boxed{\sf a=gsin\theta}}$

From 2nd Equation of motion :-

$\large{\boxed{\sf s=ut+\frac{1}{2}at^2}}$

$\large\implies{\sf 5=0+\frac{1}{2}×gsin\theta×t^2 }$

$\large\implies{\sf 5=0+\frac{1}{2}×gsin30°×t^2}$

$\large\implies{\sf t^2=\frac{10}{gsin30°} }$

$\large\implies{\sf t^2=\frac{10}{10}×\frac{2}{1} }$

$\large\implies{\sf t^2=2 }$

$\large\implies{\sf t=\sqrt{2} }$

$\large\implies{\sf t=1.41\:sec}$

$\large\red{\boxed{\sf Time(t)=1.41\:sec}}$

Time taken by the body to reach the bottom of the planet is 1.41 sec

Answer 2

$\huge{\mathfrak{\underline{\pink{Answer:-}}}}$

$\star\large\purple{\boxed{\bf Time=1.41seconds }}\star$

$\large\bf{\bold{\underline{\underline{Given:-}}}}$

$\large\star\sf\ length(s)=5m$

$\large\star\sf\ Angle \: of \: inclination (θ)=30$

$\large\star\sf\ g=10m/s^2$

$\large\star\sf\ initially \: at \: rest(u)=0$

$\large\bf{\bold{\underline{\underline{To \: find:-}}}}$

$\large\star\sf\ Time(t)=?$

$\large\sf{\bold{\underline{\underline{Solution:-}}}}$

As we know ;

$\large\purple{\boxed{\bf Accleration(a)=gsinθ}}$

Also, We use second Newton law relation:-

$\large\purple{\boxed{\bf S= ut + \frac {1}{2} at^2 }}$

let constitute the values ;

$\large\sf\implies\ 5= 0 + \frac{1}{2} \times\ gsin30° \times\ t^2 \\ \\ \large\sf\implies\ t^2= \frac {10}{gsin30°}$

$\large\sf\implies\ t ^2 = \frac {10 \times\ 2}{10 \times\ 1} \\ \\ \large\sf\implies\ t^2=2 \\ \\ \large\sf\implies\ t=\sqrt 2 \\ \\ \large\bf\implies\ t= 1.41seconds$