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Question

A daredevil jumps a canyon 13 m wide. To do so, he drives a car up a 14° incline.
(a) What minimum speed must he achieve to clear the canyon?
m/s
(b) If the daredevil jumps at this minimum speed, what will his speed be when he reaches the other side?
m/s

Answer 1

vertical velocity = -9.8t +vo sin 14º
y= -1/2*9.8*t^2 +vo sin 14 *t
x= vo cos 14º t = 0.97vo*t so t= x/0.97vo and
y=-1/2*9.8(x/0.97vo)^2+x/0.97* 0.242
y=-x^2*5.21/vo^2 +0.25x
now y=0 and x= 13
0=-169*5.21/vo2 +3.24
so vo = sqrt(169*5.21/3.24)=16.49m/s
t for y = 0 = 2 vo sin 14 /9.8 = 0.814 s
components of velocity -3.99m/s(vertical)
horizontal component = 16m/s
speed = sqrt( 256+15.90)=16,49m/s