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?
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=
2 answers
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!!
V^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g + 10 = -(5^2)/-19.6 = 11.28
m.
h = 0.5g*T^2 = 11.28 m.
4.9T^2 = 11.28.
T^2 = 2.30.
T = 1.52 s.
d = Vt*T = 30 * 1.52 = 45.6 m.
h = -Vo^2/2g + 10 = -(5^2)/-19.6 = 11.28
m.
h = 0.5g*T^2 = 11.28 m.
4.9T^2 = 11.28.
T^2 = 2.30.
T = 1.52 s.
d = Vt*T = 30 * 1.52 = 45.6 m.