Question

53. A body thrown vertically up with initial velocity
52 m/s from the ground passes twice a point at h
height above at an interval of 10 s. The height his
(g = 10 m/s2)
(1) 22 m
(2) 10.2 m
(3) 11.2 m
(4) 15 m​

Answer 1

Answer:

• The height (h) is 10 m ≈ 10.2 m.

Given:

1. Initial velocity (u) = 52 m/s.
2. Time interval (t) = 10 seconds.
3. Acceleration due to gravity (g) = - 10 m/s².

Explanation:

_______________________

From second kinematic equation,

⇒ S = u t + 1 / 2 a t²

Where,

• S denotes distance.
• u denotes initial velocity.
• t denotes time taken.
• a denotes acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ S = 52 × 10 + 1 / 2 × ( - 10) × (10)²

Here, body is thrown upward .So, acceleration due to gravity (g) will be negative.

⇒ S = 520 - 1 / 2 × 10 × 100

⇒ S = 520 - 5 × 100

⇒ S = 520 - 500

⇒ S = 20

⇒ S = 20 m.

But the question states that the body travels twice from point " h".

Therefore, Relation comes as,

⇒ S = 2 h

Where,

• S is Distance travelled in 10 seconds.
• h is the height through which ball passes twice.

Substituting,

⇒ S = 2 h

⇒ 20 = 2 h

⇒ h = 20 / 2

⇒ h = 10

⇒ h = 10 m.

∴ The height (h) is 10 m ≈ 10.2 m.

Hence, Option - (2) is correct.

_______________________