A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s in the downward direction , what will be the height attained by the stone and how much time will it take to reach there?
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Answers
Answered by
15
simple
v=0 and u is negative
⇔v=u + gt
⇔so time is 0.02 seconds
0= - 5 + (-10) * t
we know that v^2-u^2=2as or 2gh
so
0² - (- 5)² = 2 * (10) * h
⇒h = 1.25m
v=0 and u is negative
⇔v=u + gt
⇔so time is 0.02 seconds
0= - 5 + (-10) * t
we know that v^2-u^2=2as or 2gh
so
0² - (- 5)² = 2 * (10) * h
⇒h = 1.25m
keerthika1998lekha:
v = u+at 0= 5 - 10 t 10t = 5 t = 1/2 t = 0.5 sec
Answered by
10
u= - 5m/s (taken as in opposite direction)
v=0 (at max. height h)
By 1st eqn of motion
v=u + gt
0= - 5 + (-10) * t
1/50 = t
t=0.02s
By 3rd eqn of motion
v² - u² =2as [s = h]
0² - (- 5)² = 2 * (10) * h
25 = 20 *h
h = 1.25m
v=0 (at max. height h)
By 1st eqn of motion
v=u + gt
0= - 5 + (-10) * t
1/50 = t
t=0.02s
By 3rd eqn of motion
v² - u² =2as [s = h]
0² - (- 5)² = 2 * (10) * h
25 = 20 *h
h = 1.25m
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