An object is thrown upward to a certain height with a velocity of 3 m/s. The time required to reach the height at which it was thrown [Hint (g=10m/s2) (use second equation of motion)]
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u = 3m/s
v = 0m/s
h = ?
2gs = v² - u²
2×-10×s = -9
-20s = -9
s = 9/20 m
v = u + gt
0 = 3+10t
t = 3/10s
2as = v² - u²
2×-10×-9/20 = v²
-9 = v²
v = 3m/s
v = u + gt
3 = 0+10t
t = 3/10s
total time taken = 3/10 + 3/10
= 6/10
= 3/5 s = 0.6s
v = 0m/s
h = ?
2gs = v² - u²
2×-10×s = -9
-20s = -9
s = 9/20 m
v = u + gt
0 = 3+10t
t = 3/10s
2as = v² - u²
2×-10×-9/20 = v²
-9 = v²
v = 3m/s
v = u + gt
3 = 0+10t
t = 3/10s
total time taken = 3/10 + 3/10
= 6/10
= 3/5 s = 0.6s
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