Question

a stone thrown vertically upwards with a velocity of 4.9 m/s calculate the velocity with which it returns to the ground and the time taken to reach the ground

Answer 1

in upward
u=4.9
g= -9.8
v=0
so, t=v-u/g
t=1/2 sec
now in downward
t=1/2
g=9.8
u=0
so,v=u+at
v=0+9.8*1/2
v=4.9 m/s

Answer 2

Answer:

Hre initial velocity u = 4.9 m/s

Acceleration a = -9.8m/s2

When it reaches the maximum height the final velocity v = 0 m/s

We use

v2 = u2 + 2as

0 = (4.9)2 + 2 (-9.8)s

0 = 24.01 - 19.6 s

19.6 s = 24.01

s = 1.225 m

Now we also know that

s = ut + 1/2 at2

1.225 = 4.9 t - 4.9 t2

t2 - t + 0.25 = 0

t2 - t + 1/4 = 0

(t - 1/2)2 = 0

t = 0.5 s