**Stunt driver hired to do a scene where a motorcycle speeds horizontally off a 30m high cliff. There's a soft landing pad located 90m from the base. Ignore air resistance. (A) How fast must he go land on the soft padding? (B) the motorcycle has an acceleration of 5 m/s^2 on a dry road surface, he needs a certain distance from the edge to build up required speed. What is the distance?**

*My attempt: (A) Y= Yo + Vxo(T) + 1/2 (g) (T^2) ---> 0= 30 + 0 + (-5) T^2 --->T=sq.rt 6 ---> T=2.44s (we use 10 instead of 9.8)*

Xf= Xo + Vxo(T) + 1/2 (Ax) T^2 ---> 90=0 + Vxo (2.44) +0 ---> Vxo = 36.8 m/s

(B) X=Xo+VxoT + 1/2 (Ax)T ---> X-Xo= (36.8)(2.44) +0 ---> Displacement = 89.8 m

Xf= Xo + Vxo(T) + 1/2 (Ax) T^2 ---> 90=0 + Vxo (2.44) +0 ---> Vxo = 36.8 m/s

(B) X=Xo+VxoT + 1/2 (Ax)T ---> X-Xo= (36.8)(2.44) +0 ---> Displacement = 89.8 m

Did I do this correct?