NCERT CLASS IX - Science Textbook
Chapter 8 - Motion
A stone is thrown in a vertically upward direction with a velocity of5 m s - '1. If the acceleration of the stone during its motion is 10 m s - '2in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answers
Answered by
3
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using newtons equaltion we get
u = 5m/s
v=0
v=u+at'
0=-5 +(-10 t)
t= 0.5 sec
now
v² = u ² +2as
0= 25 - 20s
s=5/4 m
s=1.25 m
using newtons equaltion we get
u = 5m/s
v=0
v=u+at'
0=-5 +(-10 t)
t= 0.5 sec
now
v² = u ² +2as
0= 25 - 20s
s=5/4 m
s=1.25 m
Answered by
5
u = initial velocity = 5 m/s
a = acceleration = - 10 m/s² minus because it is in the opposite direction to velocity.
final velocity v will be 0, when the stone reaches the top point before it starts to fall back.
equation of motion: v² = u² + 2 a s
0 = 5² - 2 * 10 * s
=> s = 25 / 20 = 1.25 meters
time taken to reach the maximum height = t seconds
we know, v = u + a t
0 = 5 - 10 * t
t = 5/10 = 0.50 seconds.
a = acceleration = - 10 m/s² minus because it is in the opposite direction to velocity.
final velocity v will be 0, when the stone reaches the top point before it starts to fall back.
equation of motion: v² = u² + 2 a s
0 = 5² - 2 * 10 * s
=> s = 25 / 20 = 1.25 meters
time taken to reach the maximum height = t seconds
we know, v = u + a t
0 = 5 - 10 * t
t = 5/10 = 0.50 seconds.
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