Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?​

Answer 1

Answer:-

Given:-

Initial velocity of the stone = 40 m/s  

Final velocity of the stone = 0 m/s  

Acceleration due to gravity = −10 ms−2

Height of the stone   = ???

To find this we use, equation of motion under gravity  v²−u² = 2gs

 

Let h be the maximum height attained by the stone.

then,

⇒ v²−u² = 2gs

⇒ 02 − 402 = 2(−10)ℎ

⇒ ℎ = (40×40) / 20

= 80m  

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) = 0m.

Answer 2

Here is your answer

Maximum height reached by stone= $\bold{80m}$

Net displacement= $\bold{0m}$

Total distance covered by stone= $\bold{160m}$

$<u>EXPLANATION</u>$

Given,

u (initial velocity)= 40 m/s

accel. i.e. g= 10m/s^2

v (final velocity)= 0 m/s

Let maximum height reached by stone be h metres

Then, using the equation of motion-

${v}^{2} = {u}^{2} - 2gh$

...(for upward throw -ve sign is used), we get

${0}^{2} = {40}^{2} - 2 \times 10 \times h$

$0 = 1600 - 20h$

$20h = 1600$

$h = \frac{1600}{20}$

$h = 80$

Therefore, Maximum height reached by stone is 80m

Net displacement is 0

Reason: Stone reaches to maximum height and then falls back on earth, 8n such case their is no change in initial and final position of the stone.

Total distance covered during ascent and descent is 80m+80m=160m

HOPE IT IS USEFUL