Question

19. A particle is thrown with an initial velocity (2ỉ +3j)m/s. The horizontal range is: (g = 10 ms-2
6
5
3
2) 3m
3)
m
4)​

Answer 1

Answer:

3m is the answer if question

Answer 2

The horizontal range is 1.2m.

Given,

the initial velocity (u) = $(2 \hat i +3 \hat j)$ $ms^{-1}$

In a right angled triangle, the x and y components of the initial velocity is given by,

⇒ $u_x= 2ms^{-1}$ and,

$u_y=3ms^{-1}$

$sin\theta=\frac{3}{\sqrt{13} }$

$cos\theta=\frac{2}{\sqrt{13} }$

acceleration due to gravity (g) = 10 $ms^{-2}$

Therefore,

the magnitude of initial velocity $(|\vec u|)=\sqrt{(2)^{2}+(3)^{2}}$

                                                   $=\sqrt{13}$ $ms^{-1}$

We know that,

the horizontal range is given as,

$R=\frac{u^{2}sin2\theta}{g}$

⇒ $R=\frac{2u^{2}sin\theta cos\theta}{g}$

⇒ $R=\frac{2\times (\sqrt{13 })^{2}\times \frac{2}{\sqrt{13} } \times \frac{3}{\sqrt{13} } }{10}$

⇒ $R=\frac{12}{10}$

$R=1.2$ $m$

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