Question

A stone is thrown vertically upward with an initial velocity of 40 m per second taking G in metre per second square find the maximum height reached by the stone what is the net displacement a stone is thrown vertically upward with initial velocity 40 metre per second taking the 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

Answer 1

Answer:

80m hmax

Explanation:

usquare /2g =40square/2×10 =1600/20=80m

Answer 2

Explanation:

Given Data :

u = 40 m/s , v=0 m/s

Displacement will be zero. As it reaches again to the level from where it was thrown up.

Now,

v^2 = u^2 - 2gS

=> 0 = 1600 - 2*10*S

=> 1600/20 = S

=> 80 m = S

hence, we get height (max.) it achieved = 80m

So, we have to calculate total distance covered by stone in going up and then coming down.

S total = 2*max.height=2*80m=160m