A projectile is fired with horizontal velocity of 330m/s from the top of a cliff 80m high. How long will it strikes the level ground at the cliff? With what velocity will it strike ?
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Answer:
Since the (vertical) velocity of the ball is given by (assuming g≈ constant)
v(t)=dsdt=v0−gt,
the speed of the object when hits the ground is the velocity at time t∗, where t∗ is the value of t when s(t)=sG=0 (if your reference system is placed at s=sG=0 for the ground).
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Answer:
1320
Explanation:
R+uT
where T=underoot2H/g
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