A stone is thrown horizontally with an initial velocities of 80m/sec. Find its velocities and displacements.?

Question

A stones is thrown from the top of a building 80 m high. (a)Find its velocity after 3 sec. (b)How far is it from the ground at this time?

Answer 1

s=ut+0.5 at^2
=80*3-0.5*9.8*3^2
=240-9.41
=230.59 m it will reach ground...

Answer 2

initial velocity is 80 m/sec horizontally
initial vertical velocity=0
H=1/2gt^2
80=1/2*9.8*t^2
t=(160/9.8)^1/2=4.05 sec approx.
displacement in the horizontal direction
=80t=324 m
(a)its velocity after 3 sec=3*9.8=29.4 m/s in the vertical direction and the same 80 m/s in the horizontal direction
(b)at this point it is at a height of 80-(1/2*9.8*9)=35.9 m
the horizontal distance=240 m
so the coordinates are (240m,35.9m)