Question

A trench mortar fires a shell at an angle of 30 with the horizontal and with a speed of 400 m/s. Find its range and the maximum height it attains.

Answer 1

Answer:

The range and the maximum height attained by the body are 14.12km and 2.038km respectively.

Explanation:

Given that,

A trench mortar fires a shell at an angle of 30* with the horizontal and with a speed of 400m/s.

We shall write the given data symbolically as

$u=400m/s,\theta=30^{0}$

The range of a body projected at an angle $\theta$ with velocity $u$ is given by the relation

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

By substituting the given values we get

$R=\frac{400^2sin(2\times30^{0})}{9.81}=\frac{160000\times0.866}{9.81}=14124.26m\\\approx14.12km$

The Maximum height of a body projected at an angle $\theta$ with velocity $u$ is given by the relation

$H_{max}=\frac{u^2sin^2\theta}{2g}$

By substituting the given values we get

$H_{max}=\frac{400^2sin^230^0}{2\times9.81} \\=2038.73m=2.038km$

Therefore,

The range and the maximum height attained by the body are 14.12km and 2.038km respectively.

#SPJ3

Answer 2

Answer:

The body can go 14,124m and can climb a maximum height of 2039m

Explanation:

The horizontal and vertical movement of a projectile at time t are:

$x=V\times t\times cos\theta$

where V denotes the starting speed and the launch angle

$y=V\times t \times sin\theta-\frac{1}{2} gt^{2}$

The velocities are the displacement's times derivatives:

$Vx=Vcos\theta$

$Vx$ is constant since it is independent of $t$

$Vy=Vsin\theta-gt$

$Velocity = Vxi + Vyj$

The velocity's magnitude is

$\sqrt{(Vx^2+Vy^2)}$

at its highest point, $Vy=0=Vsin\theta-gt$

So,

at maximum height,

$t=\frac{(Vsin\theta)}{g}$

$total.flight.time T=\frac{2Vsin\theta}{g}$

The maximum height H is

$H=\frac{V^2sin^2\theta}{2g}$

A projectile thrown at an angle and travelling at velocity V has the following range R:

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

In this case,

$\theta=30$°

$V = 400m/s$

$g = 9.81m/s^2$

$so,$

$range R=\frac{400^2(0.866)}{9.81}$

             $= 14,124m$

$and$

$Height H =\frac{400^2(0.25)}{19.62}$

          $H = 2039m$

#SPJ3