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A high diver leaves the end of a 5.0 m high diving board and strikes the water 1.5 s later, 2.4 m beyond the end of the board. considering the diver as a particle, determine the initial velocity

Respuesta :

horizontal => vx = d/t = 2.4m/1.5m = 1.6

 

vertical => S = S0 + Vvt + 1/2at^2

 

0 = 5m + Vv1.3 − 4.915(1.5)^2

0 = 1.5 Vv − 6.1m

Vv = 4.6m/s

 

So, |V| = sqrt[(4.6)^2 + 1.5^2]

The calculated answer was 6.85m/s

 

The magnitude of her initial velocity is 6.85 m/s

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