A projectile is projected with initial velocity 6i 8j

Asked by admin @ 11/08/2021 in Physics viewed by 313 People

the velocity of projection of a projectile is (6i+8j)m/s.what is he horizontally range of projectile.(g=10m/s^2)

Answered by admin @ 11/08/2021

Given that,

Velocity of projection v= (6i+8j) m/s

Acceleration due to gravity = 10 m/s²

The x component of velocity is

$u_{x}=6\ m/s$

$u\cos\theta=6\ m/s$

The y component of velocity is

$u_{y}=8\ m/s$

$u\sin\theta=8\ m/s$

We need to calculate the horizontal range of projectile

Using formula range

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

$R=\dfrac{u^22\sin\theta\cos\theta}{g}$

$R=\dfrac{2(u\sin\theta)(u\cos\theta)}{g}$

Put the value into the formula

$R=\dfrac{2\times8\times6}{10}$

$R=9.6\ m$

Hence, The horizontal range of projectile is 9.6 m.

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