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I am standing next to the edge of a cliff. I throw a ball upwards and notice that 4 seconds later it is traveling downwards at 10 m/s. Where is the ball located at this time? (Pick the answer closest to the true value.)A. 120 meters above me B. 30 meters below meC. 30 meters above meD. 120 meters below meE. At the same height that it started

1 Answer

5 votes

Answer:

Step-by-step explanation:

Given

Velocity after t=4 sec is v=10 m/s downward

assuming u is the initial upward velocity


v=u+at


-10=u-gt


u=9.8* 4-10=29.2 m/s


v^2-u^2=2 as


(-10)^2-(29.2)^2=2* (-9.8)\cdot s


s=(29.2^2-10^2)/(2* 9.8)


s=38.4 m

i.e. 38.4 m above the initial thrown Position

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