Question

A stone is thrown upward with the initial velocity of 19.6 m/s. Calculate the height to which the stone will reach and after how long it will strike back to the ground.

Answer 1

g =(-9.8)m/s2

u is given

v =0 as it stops after reaching top

v2=u2+2as

0=19.6^2+(-19.6)s

s=(-19.6)^2/-19.6

s=19.6m (height)

now, u =0 as it is on top

t=?

s=

s=ut +1/2 at2

Answer 2

$\huge\mathtt{\fbox{\red{Answer}}}$

g =(-9.8)m/s2

u is given

v =0 as it stops after reaching top

v2=u2+2as

0=19.6^2+(-19.6)s

s=(-19.6)^2/-19.6

s=19.6m (height)

now, u =0 as it is on top

t=?

s=

s=ut +1/2 at^2