Question

A stone is throw straight forward with initial speed 8 m/s from a height of 25m. Find (a) the time it takes to reach the ground and (b) the speed with which it strikes.

Answer 1

Given :

Initial velocity of the stone = 8m/s

Height = 25m

To find :

(a) The time it takes to reach the ground

(b) The speed with which it strikes.

Solution :

(a) The time in which body comes down from maximum height is known as time of descent.

Time to descent is given by,

$\underline{ \boxed{\sf t = \sqrt{ \dfrac{2h}{g}}}}$

Here,

• t denotes time
• H denotes height
• g denotes acceleration due to gravity

By substituting all the given values in the formula,

$\dashrightarrow \: \sf t = \sqrt{ \dfrac{2h}{g}}$

$\dashrightarrow \: \sf t = \sqrt{ \dfrac{2(25)}{10}}$

$\dashrightarrow \: \sf t = \sqrt{ \dfrac{50}{10}}$

$\dashrightarrow \: \sf t = \sqrt{5}$

$\dashrightarrow \: \sf t = 2.23 \: sec$

Thus, the time it takes to reach the ground is 2.23 seconds.

(b) Using third equation of kinematics that is

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

where,

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height

by substituting all the given values,

$\dashrightarrow \: \sf {v}^{2} = {u}^{2} + 2gh$

$\dashrightarrow \: \sf {v}^{2} = {8}^{2} + 2(10)(25)$

$\dashrightarrow \: \sf {v}^{2} = 64 + 500$

$\dashrightarrow \: \sf {v}^{2} = 564$

$\dashrightarrow \: \sf v = \sqrt{564}$

$\dashrightarrow \: \sf v = 23.7 \: {ms}^{ - 1}$

Thus, the stone will strike with the speed of 23.7 m/s.