Question

A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

CLASS - XI PHYSICS (Kinematics)

Answer 1

Answer:

The speed of the particle at a height is 19.6 m/s.

Explanation:

Given that,

It attains a height (h) after 5 seconds and again after 9s comes back.

As the particle comes to the same point as 9s where it was at 5s. The net displacement at 4s is zero.

We need to calculate the speed of the particle at a height

Using equation of motion

$s=ut-\dfrac{1}{2}gt^2$

Put the value into the formula

$0=4v-\dfrac{1}{2}\times9.8\times(4)^2$

$4v=\dfrac{1}{2}\times9.8\times(4)^2$

$v=\dfrac{\dfrac{1}{2}\times9.8\times(4)^2}{4}$

$v=19.6\ m/s$

Hence, The speed of the particle at a height is 19.6 m/s.

Answer 2

Answer:

Medium level problem

Explanation:

It attains a height (h) after 5 seconds and again after 9s comes back.

As the particle comes to the same point as 9s where it was at 5s. The net displacement at 4s is zero.

We need to calculate the speed of the particle at a height

Using equation of motion

s=ut-\dfrac{1}{2}gt^2s=ut−

2

1

gt

2

Put the value into the formula

0=4v-\dfrac{1}{2}\times9.8\times(4)^20=4v−

2

1

×9.8×(4)

2

4v=\dfrac{1}{2}\times9.8\times(4)^24v=

2

1

×9.8×(4)

2

v=\dfrac{\dfrac{1}{2}\times9.8\times(4)^2}{4}v=

4

2

1

×9.8×(4)

2

v=19.6\ m/sv=19.6 m/s

Hence, The speed of the particle at a height is 19.6 m/s