Question

A projectile thrown from a height of 10 m with velocity √2 m/s , the projectile will fall from the foot of projection , at distance (g = 10 m/s²) :​

Answer 1

Answer:

this is the answer hope its help u

Answer 2

Given:

A projectile thrown from a height of 10 m with velocity √2 m/s.

To find:

Distance of projectile from foot of tower where it hits the ground?

Calculation:

• Let's assume that projectile is thrown horizontally from the tower.

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

$\rm \implies \: h = 0(t) + \dfrac{1}{2} g {t}^{2}$

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

$\rm \implies \: 10 = 5 {t}^{2}$

$\rm \implies \: t = \sqrt{2} \: sec$

Now, Range will be :

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

$\rm \implies\: R = \sqrt{2} \times \sqrt{2}$

$\rm \implies\: R = 2 \: metres$

So, distance from foot of tower where it hits the ground is 2 metres.