Question

A stone is thrown horizontally with a speed of
5 m/s from a building of height 5 m. What is its
displacement on striking the ground?
(1) 5 m
(2) 52 m
1910 m
14+25 m

Answer 1

Given:

A stone is thrown horizontally with a speed of 5 m/s from a building of height 5 m.

To find:

Displacement on striking the ground?

Calculation:

This is a question of Height-Ground projectile.

$\rm \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}$

$\rm \implies \: 5=0 + 5 {t}^{2}$

$\rm \implies \: {t}^{2} = 1$

$\rm \implies \: t = 1 \: sec$

Now, range of projectile is:

$\rm \: R = u_{x}(t)$

$\rm \implies\: R = 5 \times 1$

$\rm \implies\: R = 5 \: m$

Now, let displacement be d :

$\rm \: d = \sqrt{ {R}^{2} + {h}^{2} }$

$\rm \implies\: d = \sqrt{ {5}^{2} + {5}^{2} }$

$\rm \implies\: d = 5 \sqrt{2} \: m$

So, displacement is 5√2 metres.

Answer 2

Explanation:

Given:

A stone is thrown horizontally with a speed of 5 m/s from a building of height 5 m.

To find:

Displacement on striking the ground?

Calculation:

This is a question of Height-Ground projectile.

\rm \: h = u_{y}(t) + \dfrac{1}{2} g {t}^{2} h=u

y

(t)+

2

1

gt

2

\rm \implies \: 5=0 + 5 {t}^{2} ⟹5=0+5t

2

\rm \implies \: {t}^{2} = 1⟹t

2

=1

\rm \implies \: t = 1 \: sec⟹t=1sec

Now, range of projectile is:

\rm \: R = u_{x}(t) R=u

x

(t)

\rm \implies\: R = 5 \times 1⟹R=5×1

\rm \implies\: R = 5 \: m⟹R=5m

Now, let displacement be d :

\rm \: d = \sqrt{ {R}^{2} + {h}^{2} } d=

R

2

+h

2

\rm \implies\: d = \sqrt{ {5}^{2} + {5}^{2} } ⟹d=

5

2

+5

2

\rm \implies\: d = 5 \sqrt{2} \: m⟹d=5

2

m

So, displacement is 5√2 metres.