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

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