Projectile Motion Problem with stunt driver

AI Thread Summary
The discussion revolves around a stunt driver needing to calculate the speed and distance required to successfully land on a soft pad after jumping off a 30m cliff. For part (A), the driver must reach a horizontal speed of 36.8 m/s to land on the pad located 90m away, calculated using projectile motion equations. In part (B), considering an acceleration of 5 m/s², the driver needs to build up the required speed before jumping, which requires a distance of approximately 135.42 meters to reach the necessary velocity. The calculations confirm that the time in the air does not affect the distance needed to accelerate to the required speed. Overall, the analysis highlights the importance of both speed and distance in achieving a successful stunt.
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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



Did I do this correct?
 
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Question is asking "build up required speed" ... So what distance he needs to come with horizontal velocity component of 36.8 m/s (which looks good) if he has acc of 5 m/s^2 ... That is before he jumps so the time he is in the air has nothing to do here.
 
So would I use:


(Vf)^2 = (Vxo)^2 + 2(Ax) (Displacement)

0-36.8^2 = 10 (Displacement)
1354.24/10 = Displacement = 135.42 m
 
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