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=
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.
To solve this problem, we can use the equations of motion under constant acceleration. In this case, gravity will act as the only acceleration on the person.
Since we are dealing with vertical motion, we can use the following kinematic equation:
x2 = x1 + v1t + (1/2)at^2
where:
x1 is the initial position (10 m),
x2 is the final position (0 m, as the person should land in the truck bed),
v1 is the initial velocity (5 m/s),
a is the acceleration (-9.8 m/s^2, considering gravity),
t is the unknown time it takes for the person to hit the bed of the truck.
As the person jumps, he will first move upward before falling back down. At the highest point of his jump, his velocity will momentarily reach zero (v2 = 0).
Now, let's set up the equation and solve for t:
0 = 10 + 5t + (1/2)(-9.8)t^2
To solve this quadratic equation, we can rearrange it into a standard form:
(1/2)(-9.8)t^2 + 5t + 10 = 0
We can use the quadratic formula to find the values of t. The general quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the equation in standard form, a = 1/2(-9.8), b = 5, and c = 10.
Plugging in these values into the quadratic formula, we can solve for t.