Question

A stone is thrown vertically upwards with an initial velocity of 48 m/s. Find (i) the maximum

height to which it rises. (ii) the total time if takes to return to the Earth.​

Answer 1

★ Solution:

Initial Velocity ( u ) = 48 m/s

Final Velocity ( v ) = 0 m/s

Gravity in towards up = + 9.8m/s²

Gravity in towards down = - 9.8 m/s²

We know that,

$\boxed{ \boxed{ \sf{{v}^{2} - {u}^{2} = 2gh}}}$

According the question,

• The maximum height =

$\implies \: {(0)}^{2} - {(48)}^{2} = 2 \times 9.8 \times s$

$\implies \: s = \dfrac{ - (48)}{2 \times 9.8}$

$\implies \: \dfrac{2304}{19.6}$

$= 117.56m \: \approx$

∴ The maximum height = 117.5511 ≈

• The total time if takes to return to the Earth.

We know that,

$\boxed{ \boxed{ \sf{v = u + gt}}}$

According the question,

$\implies \: 0 = 48 + ( - 9.8) \times t$

$\implies \: 9.8 \times t = 48$

$= \dfrac{48}{9.8}$

$= 4.8980 \: s \: \approx$

★ Required Answer:

• The maximum height is 117.5511m ≈
• The total time if takes to return to the Earth is 4.8980s ≈