Question

a person throw a bottle into a dust bin at the same height as he is 2m away at the angle of 45 degree.the velocity of thrown is A)g B)root g C)2g D)root 2g

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Velocity=\sqrt{2g}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Range(R) = 2 \: m \\ \\ \tt: \implies Angle \: of \: projection( \theta) = 45 \degree \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Velocity \: of \: bottle = ?$

• According to given question :

$\tt \circ \: Acceleration \: due \: to \: gravity = g \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Range = \frac{ {u}^{2} sin \: 2 \theta}{g} \\ \\ \tt: \implies 2 = \frac{ {u}^{2} sin \: 2 \times 45 \degree }{g} \\ \\ \tt: \implies 2 = \frac{ {u}^{2} sin \: 90 \degree }{g} \\ \\ \tt: \implies 2g = {u}^{2} \times 1\\ \\ \tt: \implies {u}^{2} = 2g \\ \\ \green{\tt: \implies u = \sqrt{2g} } \\ \\ \green{\tt \therefore Velocity \: of \: bottle \: thrown \: is \: \sqrt{2g} }$

Answer 2

Answer is  V = $\sqrt{2g}$

Thrown at an angle of 45°

Let the initial Velocity of bottle be V

Time of Flight : $\frac{2u}{g}$ = here, $\frac{v\sqrt{2} }{g}$

Now, using $S = ut -\frac{1}{2}at^{2}$

as there is only gravity as acceleration

so no acceleration in X- Direction

So S=ut

where u is V Cos∅ where  ∅ = 45°

and t = $\frac{v\sqrt{2} }{g}$

and Dustbin is at 2m So

2 = $\frac{v}{\sqrt{2} }$ ×  $\frac{v\sqrt{2} }{g}$

So V = $\sqrt{2g}$

___________________

How Time of Flight is Derived

from $S = ut -\frac{1}{2}at^{2}$

S = 0 , so   $ut =\frac{1}{2}at^{2}$

t=2u/g