Question

The velocity of projectile is (6 i +8j) m/s. The horizontal range of the projectile is​

Answer 1

Answer:

range formula is (u²sin2Ф)/g

explanation:

tanФ=8/6=4/3

=>Ф=53 degree

velocity is 10m/s²

range=(2x10x10x4x3)/(10x5x5)

Next its up to u

solve and get the answer

hope this helps u

Answer 2

Find the magnitude of the velocity of the vector.

$\left| v \right| \\ = \sqrt{ \: {6}^{2} + {8}^{2} } \\ = \sqrt{36 + 64} \\ = \sqrt{100} = 10m {s}^{ - 1}$

Now, Given Vector is 6i +8 j.

Here, i is the unit vector along X-axis,

j is the unit vector alone Y-axis.

Consider the vector makes an angle θ with the Coordinate axes.

X, Y components are given as 6, 8 respectively.

From this,

sin2θ = 2sinθcosθ = 2(8/10)*(6/10) = 24/25.

Now,

We know that

Range of a projectile is given by,

$R = \frac{u ^{2} \sin(2 \theta) }{g} \\ \\ = \frac{ {10}^{2} \times \: \frac{24}{25} }{10} = \frac{24}{25} \times 10 \\ \\ = \frac{48}{5} \\ \\ = 9.6m$

Therefore, Range of the projectile with the velocity (6 i +8j) m/s is 9.6m