Question

An object thrown vertically upwards reached a height of 750 m. What was its initial velocity?
How long will the object take to come back to the Earth? Assume g = 10 m
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Answer 1

Given :

• Height = 750 m
• Acceleration due to gravity = 10 m/s²

To find:-

• Initial velocity = ?
• Time taken to come back to earth

Solution:-

Firstly finding initial velocity

Using the 3rd equation of motion

v² - u² = 2gs

Where

• v → final velocity = 0 m/s
• u → initial velocity = ?
• g → acceleration due to gravity = -10 m/s²
• s → distance travelled = 750m

Note ! :- Here, final velocity will be zero because the object goes up and stops & come down because of acceleration due to gravity. And acceleration due to gravity will be - ve because the object is going upward direction.

Substituting the known values we have

→ 0² - u² = 2(-10)(750)

→ 0 - u² = 20(750)

→ - u² = - 15000

→ 15000 = u²

→ u = √15000

→ u = 122.4 m/s

Now , finding time taken to come down

Using the first equation of motion

v = u + gt

Where

• v = 0 m/s
• u = 122.4 m/s
• g = - 10 m/s²
• t = ?

Substituting the values we have

→ 0 = 122.4 + (-10)×t

→ 0 = 122.4 + -10t

→ 10t = 122.4

→ t = 122.4/10

→ t = 12.24 s

Hence , time taken by the object to come down = 12.24 m/s & initial velocity of the object = 122.4 m/s.

Answer 2

Explanation:

Given that, an object thrown vertically upwards reached a height of 750 m. (At highest point final velocity = 0) and g is 10 m/s².

Using the third equation of motion,

v² - u² = 2as

Substitute the values,

→ (0)² - u² = 2(-10)(750) [g is negative as it is against the motion]

→ u² = 1500

→ u = 122.47 m/s

Hence, the initial velocity is 122.47 m/s.

Now, using the first equation of motion,

v = u + at

Substitute the values,

→ 0 = 122.47 + (-10)t

→ 122.47 = 10t

→ 12.25 = t

Hence, the time taken is 12.25 sec.