Question

An object is thrown vertically upward to a height of 10 m.Calculate its velocity with which it was
thrown upwards and time taken by it to reach a height of 10 m.​

Answer 1

Given :

• An object is thrown vertically upward to a height of 10 m.

To Find :

• The velocity with which it was
• thrown upwards = ?
• Time taken by it to reach a height of 10 m = ?

Solution :

༓ Final Velocity (v) = 0 m/s

༓ Distance travelled (s) = 10 m

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

༓ Initial Velocity (u) = ?

༓ Time taken (t) = ?

⌬ First of all we will find the initial velocity of the object by using third equation of motion :

→ v² = u² + 2as

→ (0)² = u² + 2(-9.8)(10)

→ 0² - u² = 2 × (-9.8) × 10

→ -u² = 2 × (-98)

→ -u² = -196

→ u² = 196

→ u = √196

→ u = 14 m/s

• Therefore,the velocity with which object was thrown upwards is 14 m/s.

⌬ Now, let's find the time taken by object to reach a height of 10 m by applying first equation of motion :

➻ v = u + at

➻ 0 = 14 + (-9.8)t

➻ 0 - 14 = -9.8t

➻ -14 = -9.8t

➻ 14 = 9.8t

➻ t = 14 ÷ 9.8

➻ t = 1.42 seconds

• Hence,the time taken by object to reach a height of 10 m is 1.42 seconds.

Answer 2

Given:-

• Maximum height at which the object thrown = 10m

Find:-

• Initial Velocity of object
• Time taken by stone to reach the maximum height.

Solution:-

we, know that

➤ v² - u² = 2as........【3rd eq. of motion】

➤ v² - u² = 2gs

where,

• Final Velocity, v = 0m/s
• Here, Acceleration is due to gravity, g = -9.8m/s²
• Height, s = 10m

↑ Substituting these values ↑

➨v² - u² = 2gs

➨(0)² - u² = 2(-9.8)(10)

➨0 - u² = 2(-98)

➨ -u² = -196

➨ u² = 196

➨ u = √(196)

➨ u = 14m/s

Hence, the Velocity by which an object is thrown upward is 14m/s

$\qquad$ ________________________

Now, using

➱ v = u + at..........[F1st eq. of motion]

➱ v = u + gt

where,

• Final Velocity, v = 0m/s
• Initial Velocity, u = 14m/s
• Acceleration due to gravity, g = -9.8m/s²

↑ Substituting these values ↑

➧ v = u + gt

➧ 0 = 14 + (-9.8)t

➧ -14 = -9.8t

➧ 14 = 9.8t

➧ 14/9.8 = t

➧ (14×10)/98 = t

➧ 140/98 = t

➧ 1.42sec = t

➧ t = 1.42sec

Hence, the time taken by the object to reach the maximum height is 1.42sec