Question

From a cliff of 49m high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.
PLEASE!! (Use equations of motion etc. and show the working!)............

Answer 1

for first stone 
h = 49m
g = 9.8m/s²
t = ?
u = 0
apply 
h = ut + gt²/2
49 = 0 + 9.8t²/2
t² = 10
t = √10 = 3.16s
for second stone
t = 3.16 - 1 = 2.16s
h = 49m
g = 9.8m/s²
u = ?
again apply
h = ut + gt²/2
49  = u*2.16 + 9.8*2.16²/2
49 = 2.16*u + 22.86
26.14/2.16 = u
u = 12.10m/s

 

Answer 2

$Let\ us\ take\ g\ =\ 9.8\ \frac{m}{sec}. \\ \\ 1st stone: \\ \\ \\ s = ut + \frac{1}{2} g t^2 \\ \\ 49 = 0\ t + 1/2 * 9.8 t^2 \\ \\ t = √10\ sec = 3.162\ sec \\ \\ The\ second\ stone\ traveled\ for\ 3.162\ -\ 1 \ seconds\ =\ 2.162\ sec. \\ \\ \\ 2nd stone: \\ \\ s = u t + \frac{1}{2} g t^2 \\ \\ 49 = u\ 2.162 + \frac{1}{2} * 9.8 * 2.162^2 \\ \\ Simplify, \ \ \ u\ =\ 12.07\ m /sec$