Question

A projectile is shot at an angle of `pi/6` and a velocity of `3 9 m/s`. How far away will the projectile land?

Answer 1

Here the required distance is nothing but the range of the projectile motion,which is given by the formula `R=(u^2 sin 2 theta)/g` where , `u` is the velocity of projection and `theta` is the angle of projection.

Given, `u=39 ms^-1,theta=(pi)/6`

So,putting the given values we get,

`R=134.4 m`

Answer 2

`"134.4 m"`

Explanation:

Range (`"R"`) of a projectile is given as

`"R" = ("u"^2 sin(2theta))/"g"`

`"R" = ("(39 m/s)"^2 × sin(2 xx π/6))/("9.8 m/s"^2) = "134.4 m"`