Question

A body thrown vertically upwards from the top of a building in 5s if the body is thrown vertically downwards with the same velocity it will take only 2s to reach the foot of the building calculate the height of the building and the velocity with which it was thrown​

Answer 1

Given :-

▪ A body if thrown vertically upwards from a building reaches the ground in 5 seconds. While if it is directly thrown vertically downwards from the building with the same velocity, it will take 2 seconds to reach the foot of the building (ground)

To Find :-

▪ Height of the building and velocity at which it was thrown.

Solution :-

Let the height of the building be h and velocity at which it was thrown be u.

Givent that if the body is thrown vertically downwards with the same velocity it will take 2 seconds to reach the foot of the building.

We know,

⇒ s = ut + 1/2 at²

⇒ s = 2u + 1/2×10×2² [ a = g = 10 m/s² ]

⇒ s = 2u + 20 ...(1)

Now, It is given that if the body is thrown vertically upwards it will take 5 seconds to reach the foot of the building. So, The body takes 3 seconds to reach the point from where it was thrown.

⇒ Time of flight = 2u / g

⇒ 3 = 2u / 10 [ g = 10 m/s² ]

⇒ 30 = 2u

⇒ u = 15 m/s

Substituting u = 15 in (1),

⇒ s = 2×15 + 20

⇒ s = 30 + 20

⇒ s = 50 m

Hence, The velocity at which the body was thrown is 15 m/s and the height of the building is 50 m.

Answer 2

Answer:

Given :-

▪ A body if thrown vertically upwards from a building reaches the ground in 5 seconds. While if it is directly thrown vertically downwards from the building with the same velocity, it will take 2 seconds to reach the foot of the building (ground)

To Find :-

▪ Height of the building and velocity at which it was thrown.

Solution :-

Let the height of the building be h and velocity at which it was thrown be u.

Givent that if the body is thrown vertically downwards with the same velocity it will take 2 seconds to reach the foot of the building.

We know,

⇒ s = ut + 1/2 at²

⇒ s = 2u + 1/2×10×2² [ a = g = 10 m/s² ]

⇒ s = 2u + 20 ...(1)

Now, It is given that if the body is thrown vertically upwards it will take 5 seconds to reach the foot of the building. So, The body takes 3 seconds to reach the point from where it was thrown.

⇒ Time of flight = 2u / g

⇒ 3 = 2u / 10 [ g = 10 m/s² ]

⇒ 30 = 2u

⇒ u = 15 m/s

Substituting u = 15 in (1),

⇒ s = 2×15 + 20

⇒ s = 30 + 20

⇒ s = 50 m

Hence, The velocity at which the body was thrown is 15 m/s and the height of the building is 50 m.

Explanation:

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