a stone is thrown vertically upwards with a velocity of 49 m/s square 1)how high will it rise? 2) how long will it take to reach maximum height?(g=10m/s square)
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initial vel. =49 m per sec
final velocity=0 m per sec
acceleration =- 10 m /s square
thus,By first equation of motion
v= u+at
0= 49 +(-10 ×t)
therefore
-49 = -10×t
t= -49/-10
t = 49÷10
t= 4.9 second
second ans = 4.9 sec
By third equation of motion
2 as= v square - u square
2×-10 × s= 0 - 49×49
-20s= -(50-1) (50-1)
-20 s =2500-100+1
-20 s= -2401
s= -2401÷ -20
s = 120.05 m
final velocity=0 m per sec
acceleration =- 10 m /s square
thus,By first equation of motion
v= u+at
0= 49 +(-10 ×t)
therefore
-49 = -10×t
t= -49/-10
t = 49÷10
t= 4.9 second
second ans = 4.9 sec
By third equation of motion
2 as= v square - u square
2×-10 × s= 0 - 49×49
-20s= -(50-1) (50-1)
-20 s =2500-100+1
-20 s= -2401
s= -2401÷ -20
s = 120.05 m
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