Question

Calculate its final velocity just before touching the ground.
18. A stone is thrown vertically upward with an initial velocity of
40 m/s. Taking = 10 m/s. find the maximum height reached
by the stone. What is the net displacement and the total
distance covered by the stone?​

Answer 1

Answer:

maximum height reached will be

v^2- u^2 = 2ah

-1600 = 2×(-10) ×h

h= 80 m

net displacement will be zero

total distance will be 2× max height reached

S = 2× 80. = 160m

Answer 2

STEP BY STEP EXPLANATION

As the stone is vertically thrown up,

its final velocity will be zero

i.e. v=0

Initial velocity, u=40m/s

g=-10m/s (as gravity is acting in the direction opposite to the direction of stone.)

Using 3rd equation of motion.

${v}^{2} - {u }^{2} = 2gh \\ 0 - {40}^{2} = 2 \times - 10 \times h \\ - 1600 = - 20h \\ h = \frac{1600}{20 } \\ h = 80m$

Here, the net displacement is 0 as the initial and final position of stone is same.

Total Distance=80m+80m=160 m

Hope it helps..

:)