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

Hey!!!


$given$
$g = - 10m \ {s}^{2}$
$u = 40 \frac{m}{s}$
$v = 0 \frac{m}{s}$
${v}^{2} = {u}^{2} + 2gh$
$0 = {40}^{2} + 2( - 10)(h)$
$0 = 1600 - 20h$
$- 1600 = - 20h$
$\frac{ - 1600}{ - 20} = h$
$80m = h$
Therefore, the distance traveled is 80m and displacement is zero because the stone will return back to the ground.


#BE BRAINLY

Answer 2

Dear Student,

◆ Answer -

Maximum height = 80 m

Total distance = 160 m

Displacement = 0 m

● Explanation -

Let h be maximum height reached by the stone.

v² = u² + 2as

0² = 40² + 2×(-10)×h

1600 = 20h

h = 1600/20

h = 80 m

When the stone reaches ground, it have covered 2h = 80×2 = 160 m distance. But stone returns back to original position, hence displacement is zero.

Hope this helps you..