The velocity of projection of a projectile is . $(6 i + 8 j) m / s$ What is the horizontal range of the projectile? ( $g = 10 m / s^{2}$ )

Open in App

Solution

Step1: Given data
The velocity of projection v= $(6 i + 8 j) m / s$
gravitational acceleration (g) = $10 m / s^{2}$
The velocity's x component is
$u_{x} = 6 m / s u \cos θ = 6 m / s$
The velocity's y component is
$u_{y} = 8 m / s u \sin θ = 8 m / s$
Step2:Formula Used
$R equals fraction numerator u squared sin 2 theta over denominator g end fraction$
here $R = r a n g e o f a n o b j e c t, u = i n t i a l v e l o c i t y$
$w e k n o w, \sin 2 θ = 2 \sin θ \cos θ$
Putting this value in the above equation :
$R space equals fraction numerator u squared 2 sin theta cos theta over denominator g end fraction R equals space fraction numerator 2 left parenthesis u sin theta right parenthesis left parenthesis u cos theta right parenthesis over denominator g end fraction$
Step3: Calculating the horizontal range of the projectile,
Putting the values in the above formula, we get :
$R equals fraction numerator 2 cross times 8 cross times 6 over denominator 10 end fraction R equals 9.6 space m divided by s$
Hence, The projectile's horizontal range is $9.6 m e t e r s$ .

Suggest Corrections

Similar questions

(6^i + 8^j) m s^{- 1}

(g = 10 m / s^{2})

(6 ˆ i + 8 ˆ j) m s^{- 1}

y = 12 x - \frac{3}{4} x^{2}

3 m/s

g = 10 {m/s}^{2}

\sqrt{3} / 2

u = 20 {ms}^{- 1}

g = 10 {m/s}^{2}

y = 16 x - (\frac{5}{4}) x^{2} .

g = 10 m / s^{2},

Join BYJU'S Learning Program