Question

a stone is thrown vertically upward with a speed of 15 metre per second how high the stone goes before returning back to the earth given a is equals to 9.8 metre per second square​

Answer 1

Required Concepts :

$\tt{v}^{2} = {u}^{2} + 2as.$

Given :

$\tt \vec{a} = - g = 9.8 \: m / {s}^{2} .$

$\tt\vec{ u }= + 15 \: m/ s.$

$\tt \vec{v} = 0.$

$\tt h = ?$

Solution :

$\tt {v}^{2} = {u}^{2} + 2as.$

$\implies \tt {0}^{2} = {15}^{2} + 2( - 9.8)s.$

$\implies \tt {0} = 225+ \cancel2 \frac{ - 98}{ \cancel{10}}s.$

$\tt \implies S\frac{98}{5} = 225.$

$\tt \implies S = \frac{225 \times 5}{98} .$

$\implies \tt S = 11.47 \: m.$

Therefore, the maximum height reach be height is 11. 47m.

Answer 2

We have,

Initial velocity = 15 m/s

Acceleration due to gravity = - 9.8 m/s² (Up)

Final velocity = 0 m/s (rest)

We know,

v² - u² = 2gh (3rd equation of linear kinematics)

⇒ - u² = - 2gh

⇒ h = u²/2g

⇒ h = (15 m/s)²/(2 × 9.8 m/s²)

⇒ h = (2250 m²/s²)/(2 × 98 m/s²)

⇒ h = 11.47 m (answer).