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

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!!

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?

Answer 4

To solve this problem, we can use the kinematic equations of motion. The equations we will need are:

1. x2 = x1 + v1 * t + (1/2) * a * t^2
2. v2 = v1 + a * t

In this case, the initial position (x1) is 10 meters high, the initial velocity (v1) is 5 m/s upwards, and the final velocity (v2) is 30 m/s downwards (since the truck is approaching). The acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it acts downwards). We want to find the distance (x2) at which the person can jump, so we need to solve for x2.

Now, let's break down the steps:

1. First, let's find the time it takes for the person to reach the truck. We can use equation (2) to solve for time (t).
v2 = v1 + a * t
30 = 5 + (-9.8) * t
9.8t = -25
t = -25 / 9.8 ≈ -2.55 (eliminating negative value)

The time it takes for the person to reach the truck is approximately 2.55 seconds.

2. Now that we have the time, we can use equation (1) to solve for x2.
x2 = x1 + v1 * t + (1/2) * a * t^2
x2 = 10 + 5 * 2.55 + (1/2) * (-9.8) * (2.55)^2
x2 ≈ 10 + 12.75 - 32.0185
x2 ≈ -9.2685

The negative result means that the person cannot jump far enough to reach the truck.
The person should wait for the next truck or make sure to jump with a higher speed to clear the bridge railing.

In conclusion, the person should wait for the next truck or jump with a higher initial velocity to clear the bridge railing.