A person is standing on a 10-m bridge above a road. He wants to jump from the bridge and land in the bed of a truck that is approaching him at 30 m/s. In order to clear the bridge railing, he has to jump upward initially with a speed of 5 m/s. How far away should the truck be when he jumps in order for him to land in the bed?

Question

What would this be and what equation should I use to solve this??
INITIAL POSITION x1=
FINAL POSITION x2=
INITIAL VELOCITY v1=
FINAL VELOCITY v2=
ACCELERATION a=
TIME t=

I have to submit before 6 pm 12/17/15. I am super confused and I have another test. Someone help show the steps and simplify what to do. Much appreciated!!

Answer 1

He is at 10 meters and must jump with an initial velocity up of 5 m/s
ow long until he reaches 0 meters (ground)

h = Hi + Vi t - 4.9 t^2

0 = 10 + 5 t - 4.9 t^2
or
4.9 t^2 - 5 t - 10 = 0
solve quadratic for time in the air, t in seconds

now the truck has to be t seconds away from the bridge.
d = 30 t

Answer 2

INITIAL POSITION x1= 10m
FINAL POSITION x2= 0 m
INITIAL VELOCITY v1= 5 m/s
FINAL VELOCITY v2=?
ACCELERATION a= -9.8 m/s^2
TIME t=?

Are these numbers put in the proper place? I'm confused on this part before I get towards the end..

Answer 3

So is T= 2.03 and is the Distance/ Final velocity= 60.9 m/s ???
So to answer the question: How far away should the truck be when he jumps in order for him to land in the bed?...

So.. the answer in terms of how far away the truck should be is 60.9 meters/ second away it is?