Question

A stone is thrown vertically upward with an initial velocity of 40 m per second taking G is equals to 10 metre per second square find the maximum height reached by the stone what is the net displacement and the total distance covered by the stone?

Answer 1

Explanation:

initial velocity = 40

final = 0

g= 10

using ,

v^2 - u^2 = 2gh

0^2- 40^2 = 2× -10 ×h

1600 = 20 ×h

h = 1600÷20

h = 80 meter

distance = displacement

therefore , distance = 80m

displacement = 80 m

Answer 2

Solution♡

According to the third equation of motion under gravity:

• v²-v²=2gh

Where,

• u= Initial velocity of the stone
• v= final velocity of the stone
• h= Displacement of the stone
• g= Acceleration due to gravity

We know that velocity at the highest point is equal to zero.

Here,

u=40 m/s

v=0

g=-10 ms^-2

Therefore,

0-(40)²=2×(-10)×h

h=40×40/20=80m

Maximum height reached=80m

Therefore,

Total distance covered by the stone during its upward and downward generally

= 18 + 18 =160m.

Net displacement of the stone during its upward and downward journey

= 80m+(-80m)=0.

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