Projectile at two instants of time

Let the horizontal and vertical components of the initial velocity $v$ be $v_x$ and $v_y=v_y(0)$ respectively. $v_x$ is constant, while $v_y(t)$ changes with time $t$ .

When the projectile is moving horizontally at $\space t=3\quad \Rightarrow v_y(3) = 0$ , and $v_y(t)$ is given by:

$v_y(t) = v_y(0) - gt\quad \Rightarrow v_y(3) = v_y - 10(3) = 0 \quad \Rightarrow v_y = 30\space m/s$

At $\space t=2\quad \Rightarrow v_y(2) = 30 - 10(2) = 10 \space m/s$

We know that: $\space \dfrac {v_y(2)}{v_x} = \tan {30^\circ} \quad \Rightarrow v_x = \dfrac {v_y(2)} {\tan {30^\circ}} = 10\sqrt{3}\space m/s$

Therefore,

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{900+300} = \sqrt{1200} = 20\sqrt{3} = 34.64101615\space m/s$

$\Rightarrow \lfloor v \rfloor = \boxed{34}$

We know Time of Flight (T) $= \frac{2u \sin x}{g}$ where $x$ is the initial angle of projection.
Here $3 = \frac{u \sin x}{g}$ , $u \sin x = 3g$

After 2 seconds,
velocity in vertical direction = $u\sin x - 2g$
velocity in horizontal direction = $u \cos x$

A/q
$\tan 30 = \frac{1}{ \sqrt{3} } = \frac {u\sin x - 2g}{u\cos x}$
$\frac{1}{ \sqrt{3} } = \frac {g}{u\cos x}$
$u \cos x = \sqrt{3}g$

So, $\tan x = \sqrt{3}$ , $x = 60$
$u = \frac{3g}{\sin 60} = 20 \sqrt{3} = 34.641016151$

1.Using the expression of time of flight over half the range, first find initial velocity along the vertical.

2.Then by using first eqn. of motion, find final velocity along the vertical.

3.Now find final velocity along the horizontal using tan 30=v along y/v along x

4.as motion along horizontal is uniform, initial velocity along horizontal is same as final velocity along horizontal

5.We now have initial velocities along both horizontal and vertical, hence,we can now find the net velocity by taking underoot of squared sum of both of them.We get 34.64 as the answer.

To be clear from the beginning:

• t = 0, projectile was fired

• t = 2, projectile is moving at 30 degrees to horizontal

• t = 3, projectile is horizontal

As g = 10, the vertical velocity will decrease by 10 m/s each second. Thus at t = 2, it must have been 10. At t = 0, it must have been 30.

The only unknown now is the horizontal velocity which is a constant throughout. At t = 2, vertical velocity is 10. Projectile is moving at an angle 30 degrees which means horizontal velocity is 10 x tan(60) = 10 x sqrt(3).

The required answer is then sqrt(300 + 900)