Question

. Thief A
is in an apartment 10m high and is throwing a bag of valuables he
high and is throwing a bag of valuables horizontally at 5m/s. Thief B is
Initially standing right below the point of release. Find the velocity with which he should run to
catch the bag in time. Give reason to support your answer.
AR​

Answer 1

Answer:

Given:

Thief A is standing at a height of 10 m.

The initial Velocity of the projectile is

5 m/s.

To find:

Speed with which the Thief B should run in order to catch the projectile.

Calculation:

For any projectile, we can divide the motion along x and y axes.

So, time to reach the ground can be calculated along y axis:

y = ut + ½at²

=> 10 = (0 × t) + ½ × 10 × t²

=> t² = 2

=> t = √2 seconds.

So range of projectile is given as :

= (horizontal velocity × time)

= 5 × √2

= 5√2 metres.

Now, thief B also starts running and has to reach the box by √2 seconds.

So, Velocity of thief B

= (5√2)/√2

= 5 m/s

Proof that trajectory is parabolic:

Distance travelled along y axis:

y = (0 × t) + ½gt² .......(1)

Distance along x axis:

x = ut .......(2)

Putting value of t in eq.(1)

y = ½g(x/u)²

=> y = ½gx²/u²

=> y = (g/2u²) x²

∴ y ∝ x² , hence parabolic.

Answer 2

Answer:

• The thief Should run with a Velocity (V) of 5 m/s.

Given:

1. Height of Apartment (y) = 10 m.
2. Horizontal velocity (u) = 5 m/s.

Explanation:

$\rule{300}{1.5}$

Applying Second kinematic equation in y - Direction.

$\large \bigstar \: {\boxed{\tt y = vt + \dfrac{1}{2}gt^2}}$

$\bold{Here}\begin{cases}\text{y Denotes Height of Apartment} \\ \text{v Denotes vertical velocity} \\ \text{g Denotes Acceleration due to gravity} \\ \text{t Denotes Time period}\end{cases}$

Now,

$\large{\boxed{\tt y = vt + \dfrac{1}{2}gt^2}}$

Vertical component of velocity will be Zero as body is projected horizontally.

Substituting,

$\longmapsto{\large{\tt y = 0 \times t + \dfrac{1}{2} \times 10 \times t^2}}$

$\longmapsto{\large{\tt y = 0 + \dfrac{1}{2} \times 10 \times t^2}}$

$\longmapsto{\large{\tt y = \dfrac{1}{2} \times 10 \times t^2}}$

$\longmapsto{\large{\tt y = \dfrac{10}{2}\times t^2}}$

$\longmapsto{\large{\tt y = \cancel{\dfrac{10}{2}}\times t^2}}$

• y = 10 m (Apartment Height)

$\longmapsto{\large{\tt 10 = 5 \times t^2}}$

$\longmapsto{\large{\tt t^2 = \dfrac{10}{5}}}$

$\longmapsto{\large{\tt t^2 = \cancel{\dfrac{10}{5}}}}$

$\longmapsto{\large{\tt t^2 = 2 }}$

$\longmapsto{\large{\underline{\boxed{\tt t = \sqrt{2} \: Seconds}}}}$

From Range Formula,

$\large \bigstar \: {\boxed{\tt R = u \times t}}$

$\bold{Here}\begin{cases}\text{R Denotes Range} \\ \text{u Denotes Horizontal velocity} \\ \text{t Denotes Time period}\end{cases}$

Now,

$\large{\boxed{\tt R = u \times t}}$

Substituting the values,

$\longmapsto \large{\tt R = 5 \times \sqrt{2}}$

$\longmapsto \large{\underline{\boxed{\tt R = 5 \sqrt{2} \: m}}}$

For Thief B

$\large \bigstar \: {\boxed{\tt V = \dfrac{R}{t}}}$

General Expression of Distance, Velocity & time.

Substituting the values,

$\longmapsto \large{\tt V = \dfrac{5 \sqrt{2}}{\sqrt{2}}}$

$\longmapsto \large{\tt V = \cancel{\dfrac{5 \sqrt{2}}{\sqrt{2}}}}$

$\longmapsto \large{\underline{\boxed{\red{\tt t = 5 \: Second}}}}$

∴ The thief Should run with a Velocity (V) of 5 m/s.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From y - Component of Displacement we have.

$\longmapsto \large{\tt y = \dfrac{1}{2}gt^2}$

And, From x - component of Displacement we have.

$\longmapsto \large{\tt x = ut}$

This can be written as,

$\longmapsto \large{\tt t = \dfrac{x}{u}}$

Substituting this Equation in y - Component of Displacement.

$\longmapsto \large{\tt y = \dfrac{1}{2}gt^2}$

$\longmapsto \large{\tt y = \dfrac{1}{2}g \times \Bigg(\dfrac{x}{u}\Bigg)^2}$

$\longmapsto \large{\tt y = \dfrac{1}{2}g \times \dfrac{x^2}{u^2}}$

Here we can see y is proportional to Square of x - component displacement.

I.e.

$\longmapsto \large{\underline{\boxed{\red{\tt y \propto x^2}}}}$

∴ The Motion or Path is Parabolic.

$\rule{300}{1.5}$