a ball is thrown vertically upward with an initial velocity of 49 m p s . calculate a) maximum height attained b) time taken by it before it reaches the ground again ( g = 9.8 m p s square )
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intial velocity = u = 49
final velocity = v = 0
acceleration = a = 9.8
s = height
v²-u² = 2 a s
intial velocity = u = 49
final velocity = v = 0
acceleration = a = 9.8
s = height
v²-u² = 2 a s
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Final velocity(v)=49 metre per second
initial velocity(u)=0
g=9.8 metre per second²
h=?
v²=u²+2gh
⇒v²=2gh
⇒h=v²÷2g
⇒h=49×49/9.8×2
⇒h=122.4999999 metres
Hence, the maximum height reached will be 122.4999999 metres.
initial velocity(u)=0
g=9.8 metre per second²
h=?
v²=u²+2gh
⇒v²=2gh
⇒h=v²÷2g
⇒h=49×49/9.8×2
⇒h=122.4999999 metres
Hence, the maximum height reached will be 122.4999999 metres.
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