Question

A stone is thrown vertically upwards with a velocity of 40m/s and is caught back. Taking g=10m/s sq. calculate the maximum height reached by stone
what is the total distance and displacement covered by stone

Answer 1

Hello,

u = 40 m/s.
g = 10 m/s

$h = \frac{u^{2} \sin^{2} ( \alpha )}{2g}$
h = [40 × 40 × sin(90) × sin(90)]/(2 × 10)
= (1600 × 1 × 1)/(20)
As sin(90) = 1
= 1600/20
= 160/2
Therefore maximum height = 80 m.

As t = (u)/(g)
= 40/10
Therefore time = 4 seconds to reach the maximum height. (neglecting air resistance)

As s = (ut) + (½ × at²)
= (ut) - (½ × gt²)
= (40 × 4) - (½ × 10 × 4 × 4)
= 160 - 80
Therefore distance covered till the maximum height is 80 m. (going up)
Therefore the total distance = distance went up + distance went down.
Therefore the total distance till the ball is caught = 80 + 80 = 160 m.

Displacement = 0 m. As there is no change between initial point and the final point.

Hope this answer helps you
If you need more help, Contact me
@YHarith #YHarith

Answer 2

_/\_Hello mate__here is your answer--

____________________

u = 40 m/s

v = 0 m/s

s = Height of the stone

g = −10 ms−2 ( upward direction)

Let h be the maximum height attained by the stone.

Using equation of motion under gravity

v^2 − u^2 = 2gs

⇒0^2 − 40^2 = 2(−10)ℎ

⇒ ℎ =40×40/ 20 = 80

Therefore, total distance covered by the stone during its upward and downward journey

= 80 + 80 = 160 m

Net displacement during its upward and downward journey

= 80 + (−80) = 0.

I hope, this will help you.☺

Thank you______❤

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