physics question?????

P

ping pong

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A diver jumps from a 3.0m board (the height of the board above the water) with an initial upward velocity of 4.0ms. Taking the acceleration of gravity to be 9.8 calculate:

a the maximum height of the divers jump

b the diver's velocity at the water surface

C the time for which the diver is in the air



ive got the answers to a and b correct but keep getting c wrong im getting 0.87 but the answer sheet says 1.3 secs how?i got 0.40 secs for going up and .47 for going down
 
use the suvat equations

a) u=4, s=?+3 a=-9.81, v=0. (vsquared = usquared + 2as)
s= 0.8 add 3 = 3.8m. max height = 3.8m

b) a=9.81, s=3.8, u=0, v=? (vsquared = usquared + 2as)
v=8.6. divers velocity at water = 8.63

c) first part of jorney -

second part of journey -
 
use the suvat equations

a) u=4, s=?+3 a=-9.81, v=0. (vsquared = usquared + 2as)
s= 0.8 add 3 = 3.8m. max height = 3.8m

b) a=9.81, s=3.8, u=0, v=? (vsquared = usquared + 2as)
v=8.6. divers velocity at water = 8.63

c) first part of jorney -

second part of journey -
 
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