Question

An object is thrown vertically upwards and rises to a height of 13.07m. Calculate
(i) the velocity with which the object was thrown upwards. (ii) the time taken by the object to reach the highest point. { take s=h(height) and a=g (acceleration due to gravity)} g=9.8m/s2​

Answer 1

Given:-

• Final velocity ,V = 0m/s

• Height ,h = 13.07 m

• Acceleration due to Gravity ,g = 9.8m/s²

To Find:-

(i) the velocity with which the object was thrown upwards. (ii) the time taken by the object to reach the highest point.

Solution:-

[i]

Firstly we calculate the initial velocity of the object. Using Third Equation of Motion

• v² = u² +2gh

v is the final velocity

g is the acceleration due to gravity

u is the initial velocity

h is the height attained by object.

Substitute the value we get

→ 0² = u² + 2 × (-9.8) × 13.07

→ 0 = u² + (-19.6) × 13.07

→ -u² = -19.6×13.07

→ -u² = -256.172

→ u² = 256.172

→ u = √256.172

→ u = 16.005 ≈ 16 m/s

Therefore, the velocity with which the object was thrown upward is 16 m/s.

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[ii]

Now, calculating the time taken to reach the height point.

Using 1st Equation of Motion

• v = u +at

Substitute the value we get

→ 0 = 16 × (-9.8) × t

→ -16 = -9.8×t

→ t = -16/-9.8

→ t = 16/9.8

→ t = 1.63 s

Therefore, the time taken by the object to reach the maximum height is 1.63 second .

Answer 2

☆Given Question:-

An object is thrown vertically upwards and rises to a height of 13.07m. Calculate

(i) the velocity with which the object was thrown upwards. (ii) the time taken by the object to reach the highest point.

{ take s=h(height) and a=g (acceleration due to gravity)} g=9.8m/s2

☆Required Answer:-

$\large \sf {Given:- }$

• An object is thrown vertically upwards and rises to a height of 13.07m.

$\large \sf {To \: Find:- }$

1. The velocity with which the object was thrown upwards.
2. The time taken by the object to reach the highest point.

$\large \sf {Dealing:- }$

• We will find the answer in two passage.

Passage 1:-

✽According to Question, The velocity by which the object was thrown upwards will be presented as 'Initial velocity'.

✽We can find the initial velocity of the object by using the third equation of motion, i.e.

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

✽As given in the Question, we will substitute the values...

$\implies \sf{ {0}^{2} - {u}^{2} = 2 \times - 9.8 \times 13.07}$

$\implies \sf{ {0} - {u}^{2} = - 19.6 \times 13.07}$

$\implies \sf{ - {u}^{2} = - 256.172}$

$\implies \sf{ {u}^{2} = 256.172}$

$\implies \sf{ {u} = \sqrt{ 256.172}}$

$\implies \sf{ {u} = 16 {ms}^{ - 1} }$

So, the initial velocity of the object is 16m/s.

Passage 2:-

✽We will use the 1st equation for motion, i.e.

$\hookrightarrow \sf{ {v} = {u} + {at}}$

✽As given in the Question, we will substitute the values...

$\implies \sf{ {0} = { 16} - 9.8 \times t}$

$\implies\sf{ t = \frac{ - 16}{ - 9.8} }$

$\implies\sf{ t = 1.63s}$

So, the time taken will be 1.63s.

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Know More:-

• v= Final velocity

• u=Initial velocity

• s= displacement

• t= time

• a=acceleration

• g=Acceleration due to gravity.

• h= height

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