i'm confused, can someone clarify?

Question

a diver runs horizontally off the end of a 5 m high drivng board at 6 m/s.
a) how long does it take for diver to hit water? is it 1.445 s?
b) how far horizontally does the diver end up from the board? i'm confused, isn't it 5 m?
c) what is size of horizontal component of the driver's velocity just before he hits the water? do I have enough info to construct one? how should it look like?
d) size of vertical y component velocity before he hits water?
e) speed in m/s? i guess size does not have direction, but how can it be differnt from its velocity?

Answer 1

a) X = Xo + VoT + ½AT²
X = 0 (the water's surface)
Xo = 5 m
Vo = 0 m/s (leaves the board horizontally)
A = -9.8 m/s² (acceleration of gravity)
Horizontal component of velocity plays no part.
Solve for T:
0 = 5 + 0T + ½(-9.8)T²
5 = 9.8T²/2
10/9.8 = T²
T = 1.01 seconds

b) The diver travels for 1.01 seconds at 6 m/s so he travels 6.06 meters horizontally.

c) The horizontal component of the diver's speed is the same the entire time - 6 m/s. It does not change as he falls.

d) Vertical velocity: V = Vo + AT
V = 0 + (-9.8)(1.01) = -9.9 m/s
minus sign means that the velocity is directed down, towards the water.

e) Speed is vector sum of horizontal component (6 m/s) and vertical component (9.9 m/s), which would be

V = sqrt(6² + 9.9²) = 11.6 m/s

Answer 2

a. x = 1/2at^2
....t = (2*9.8/5)^1.2 =2.0 sec (to 2 sig figs)

b. x = vt = 6*2 = 12m (to 2 sig figs)

c. 6 m/s, it does not change

d. v = at = 9.8 * 2 = 18 m/s (to 2 sig figs)

e. Resolve the vectors
...S = (6^2 + 18^2)^1/2 = 19 m/s (to 2 sig figs)