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A physics student falls from 12.0 cliff How long does it take him to reach the ground and what is his velocity by the time he reaches the ground? |
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Answer» ong>Answer: Height of the cliff h 1
=20m Velocity of projectile V=30m/s Initial upward velocity v y
=Vsinθ=30×sin30 ∘
=15m/s. HORIZONTAL velocity, v x
=Vcosθ=30cos30 o m/s Time taken to REACH maximum height, t 1
= g V y
= 10 15
=1.5sec. Height above the cliff h 2
= 2g v y 2
= 2×10 15 2
=11.25m Therefore, the maximum height, h max
=h 1
+h 2
=20+11.25=31.25m. Time taken to free fall from max height, t 2
= g 2h max
= g 2×31.25
=2.5sec. Thus, the total time taken during the entire flight t total
=t 1
+t 2
=1.5+2.5=4sec Total horizontal distance covered R=v x
t total
=30×cos30 ∘
×4=60 3
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