Question

When a particle is thrown vertically upwards, its velocity at one third of its maximum height is 10√2

m/s . The maximum height attained by it is

(1) 20√2 m

(2) 30 m

(3) 15 m

(4) 12.8 m

Answer 1

Answer:

The maximum height attained by it is 15 m

Explanation:

Given that,

Velocity $v=10\sqrt{3}$

Let h be the maximum height reached.

The velocity of the particle when it was reached its maximum height is zero.

Distance traveled is

$h-\dfrac{h}{3}=\dfrac{2h}{3}$

Using equation of motion

$v^2=u^2+2gh$

Where, v = velocity

g = acceleration due to gravity

h = height

Put the value in equation

$(10\sqrt{2})^2=2\times9.8\times\dfrac{2h}{3}$

$h=\dfrac{3\times(10\sqrt{2})^2}{4\times9.8}$

$h = 15.3\ m$

$h = 15\ m$

Hence, The maximum height attained by it is 15 m

Answer 2

Explanation:

The maximum height attained by it is 15 m

Explanation:

Given that,

Velocity v=10\sqrt{3}v=10

3

Let h be the maximum height reached.

The velocity of the particle when it was reached its maximum height is zero.

Distance traveled is

h-\dfrac{h}{3}=\dfrac{2h}{3}h−

3

h

=

3

2h

Using equation of motion

v^2=u^2+2ghv

2

=u

2

+2gh

Where, v = velocity

g = acceleration due to gravity

h = height

Put the value in equation

(10\sqrt{2})^2=2\times9.8\times\dfrac{2h}{3}(10

2

)

2

=2×9.8×

3

2h

h=\dfrac{3\times(10\sqrt{2})^2}{4\times9.8}h=

4×9.8

3×(10

2

)

2

h = 15.3\ mh=15.3 m

h = 15\ mh=15 m

Hence, The maximum height attained by it is 15 m