Question

Two children P and Q of equal masses are at a swimming bath . Child P drops vertically from a diving board 5m high. Child Q slides from the same height (5m) down a slide into the water. The children start moving at the same instant . The children start moving at same instant . (Ignore air resistance and friction on slide).which of the following is true?
1 )the children hit the water same time at same speed
2) they hit the water at different times with different speed
3)they hit the water at different speed but same time
4)they hit the water at different time but same speed
5) none of these


*** Please respond only if you know the answer.

Answer 1

Answer:

4)they hit the water at different time but same speed

Explanation:

As we know that one child jump into the water and fall freely into it

So here we know that

$h = \frac{1}{2} gt^2$

now we have

$t = \sqrt{\frac{2h}{g}}$

now his speed is given as

$v = \sqrt{2gh}$

now for other child on inclined plane we have

$L = \frac{1}{2}(gsin\theta) t^2$

so we have

$t = \sqrt{\frac{2L}{g sin\theta}}$

here we can find the speed as

$v = \sqrt{2(gsin\theta)L}$

as we know that

$Lsin\theta = h$

so we have

$v = \sqrt{2gh}$

now correct answer is

4)they hit the water at different time but same speed

#LEarn

Topic : kinematics

https://brainly.in/question/5536466

Answer 2

Answer:

4)they hit the water at different time but same speed

Explanation:

As we know that one child jump into the water and fall freely into it

So here we know that

h = \frac{1}{2} gt^2h=

2

1

gt

2

now we have

t = \sqrt{\frac{2h}{g}}t=

g

2h

now his speed is given as

v = \sqrt{2gh}v=

2gh

now for other child on inclined plane we have

L = \frac{1}{2}(gsin\theta) t^2L=

2

1

(gsinθ)t

2

so we have

t = \sqrt{\frac{2L}{g sin\theta}}t=

gsinθ

2L

here we can find the speed as

v = \sqrt{2(gsin\theta)L}v=

2(gsinθ)L

as we know that

Lsin\theta = hLsinθ=h

so we have

v = \sqrt{2gh}v=

2gh

now correct answer is

4)they hit the water at different time but same speed