Question

Ahoy all

A stone is thrown vertically upward with a velocity of 40 m per second and returns back. Taking g = 10 m/s² , calculate the maximum height reached by the stone and the total distance travelled by the stone.

solve it!

Answer 1

Question :


A stone is thrown vertically upward with a velocity of 40 m per second and returns back. Taking g = 10 m/s² , calculate the maximum height reached by the stone and the total distance travelled by the stone.



Solution :


Initial velocity = u = 40 m/s

Final velocity = v = 0

A = -g = -10 m/s²

By using v²-u² = 2as

0²-40²=2×-10×h

-1600 = -20h

h = 1600/20 = 80 m

Total Distance Covered = 2h = 2×80m = 160 m

Answer 2

Here is your answer mate ↙️↙️✌️✌️✅✅

According to the equation of motion under gravity v2 − u2 = 2gs 

Where, 

u = Initial velocity of the stone = 40 m/s 

v = Final velocity of the stone = 0 m/s 

s = Height of the stone 

g = Acceleration due to gravity = −10 ms−2 

Let h be the maximum height attained by the stone. 

Therefore, 0 2 − 402 = 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.

Hope this helps you dude all the best dude
☺️☺️✌️✌️✌️✅✅✅✅