Question

A Stone thrown from the top of a building is given an initial velocity Of 20.0 metre per second straight upward determine the time in second at which the stone reaches its maximum height. ( Take sec2 g=9.8m/s)
A) 2.8
B) 2.04
C) 1.67
D) 2.7​

Answer 1

Answer:

B) 2.04

Explanation:

Given that,

• Initial velocity of the stone (u) = 20.0 m/s
• Acceleration due to gravity (g)= -9.8 m/s² [upward motion]
• Final velocity of the stone $\sf{(v_{max})=0\:m/s}$

We know that,

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

• [ Put values ]

$\implies \sf \: {0} \: = {20} + ( - 9.8) \times t \\ \\ \implies \sf \: 0 = 20 - 9.8t \\ \\ \implies \sf \: 9.8t = 20 \\ \\ \implies \sf \: t = \dfrac{20}{9.8} \\ \\ \implies \sf \: t = 2.04$

Therefore, the time at which the stone reaches its maximum height will be 2.04 s.

Answer 2

Given :-

• Initial Velocity (u) = 20m/s
• Final Velocity (v) = 0
• Acceleration (a) = -9.8 m/s²  [Retardation OR Deceleration is taking place so acceleration is negative]

To Find :-

• Time taken to reach the maximum height = ?

Concept Implemented :-

We need to use the Equation of Motion

❶ First Equation of Motion

v = u + at

❷ Second Equation of Motion

s = ut + $\frac{1}{2}$at²

❸ Third Equation of Motion

v² = 2as + u²

Diagram :-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines \put(0,0){\framebox(4,4)}\put(1.4,2){\bf{Building}} \put(1.9,4.2){\circle*{.5}} \put(1.9,4){\vector(0,1){3}} \put(1.9,7.3){\circle*{.5}} \put(2.4,7.3){\bf{v = 0}}\put(2.4,4.2){\bf{u = 20 m/s}}\qbezier(0,7.6)(0,7.6)(4,7.6)\put(.6,7.9){\bf{Maximum Height}}\end{picture}$

● NOTE : If diagram isn't visible then refer to the attached image.

Solution :-

We will First Equation of Motion

◆ v = u + at

✵ Putting the values in the Equation ✵

0 = 20 + [(-9.8) × t]

⇒ 0 = 20 - 9.8t

⇒ 9.8t = 20

$\sf \implies t = \dfrac{20}{9.8}$

⇒ t = 2.04 seconds

∴ The stone reaches it's maximum height in 2.04 seconds.