Question

Question 15 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Class 9 - Science - Gravitation Page 144

Answer 1

the maximum height reached is 50.2

Answer 2

Here, u = 40 m/s,

g = - 10 m/s²

v = 0 (at the maximum height)

S = ?


◆ ¡) Using, 2aS = v²- u², we get

S = v²- u²/ 2a

= 0 - 40 × 40/ -2 × 10

= 80 m

Therefore, Maximum height = 80 m.


◆ ¡¡) The stone after reaching the maximum height falls down and reach the ground.

Therefore, Displacement of the stone = 0

and , Distance covered by the stone
= 80m + 80m

= 160 m.