A daredevil performs a stunt in which he jumps with his motorcycle from a horizontal cliff 10m high. What must the initial velocity of the motorcycle be, in order for the stuntman to barely clear a 10m wide water canal at the base of the cliff?

Answer 1

distance across=vhorizongal*time

now, in vertical, d=1/2 g t^2 OR
t=sqrt(20/9.8)

put that time in the first equatioan, solve for vhorizontal

Answer 2

I got the wrong answer. My answer key says the answer is 7m/s.

t = sqrt(20/9.8) = 1.428

d = 1/2(9.8)(1.428)^2 = 10

d*t = 14.28

What am I doing wrong?

Answer 3

h = 0.5g*t^2 = 10 m.
h = 4.9*t^2 = 10
t^2 = 2.04
Tf = 1.43 s

Xo*1.43 = 10 m.
Xo = 7.0 m/s