Question : A person standing on a tower simultaneously launches two apples, one straight up with a speed of 7 m/s and the other straight down at the same speed. How far apart (in m) are the two apples 2 seconds after they were thrown, assuming that neither has hit the ground?
Doubt by Suhani
Solution :
Case I
The apple which is launched upwards
initial velocity (u) = 7 m/s
acceleration (a) = -g
initial velocity (u) = 7 m/s
acceleration (a) = -g
time (t) = 2sec
Distance covered (s1)
Using II Equation of Motion
s=ut+½at²
s1=ut+½(-g)t²
s1=ut-gt²/2 — (1)
s1=ut-gt²/2 — (1)
Case II
The apple which is launched downwards
initial velocity (u) = 7 m/s
acceleration (a) = +g
initial velocity (u) = 7 m/s
acceleration (a) = +g
time (t) = 2sec
Distance covered (s2)
Using II Equation of Motion
s=ut+½at²
s2=ut+½(+g)t²
s2=ut+gt²/2 — (2)
s2=ut+gt²/2 — (2)
Distance between the two apples after 2 seconds
= s1+s2
= ut-gt²/2 + ut+gt²/2
=ut+ut
= s1+s2
= ut-gt²/2 + ut+gt²/2
=ut+ut
=2ut
=2(7)(2)
=2(7)(2)
= 28 m
Hence, the two apples are 28 m far apart after 2 seconds of launching.