Abungee jumper falls from a bridge 400 feet above ground. her height above ground after t seconds is given by the equation where h is the height of the jumper in feet and t is the time in seconds since her jump. approximately how long will it take for the bungee jumper to reach 100 feet above the ground on her initial jump?
B. 4.3 seconds.
1. You must substitute h=100 into the equation given in the problem and solve for the time, as you can see below:
3. Then, it will take 4.3 seconds for the bungee jumper to reach 100 feet above the ground on her initial jump. Therefore, the answer is the option B.
A) The time it takes for the T-shirt to reach the fan is 2.3582s
B) The height the fan is from the ground is 10.2438m
First, we need to find the velocity vectors for each axe:
Next, using the movement equation for the x axis, we get the time:
Finally, using the movement equation for the y axis, we find the height (assuming the gravity as 9,8)
B. 4.3 seconds
Given: The height above ground after t seconds is given by the equation h=-16t^2+400 where h is the height of the jumper in feet and t is the time in seconds since her jump.
To find time to reach the height (h) = 100 ft above the ground, we need to plug h = 100 in the given equation and solve for t.
100 = -16t^2 + 400
Subtract 400 from both sides, we get
100 - 400 = -16t^2 + 400 - 400
-300 = -16t^2
Now divide both sides by -16, we get
-300/-16 = -16t^2/-16
t^2 = 18.75
Now take square root on both sides.
√t^2 = √18.75
t = 4.3 seconds.
The police car is moving at a constant speed
Since the car is moving at constant speed then, it acceleration is zero.
Another car who is at a high speed passed the police at a speed of
V' = 145km/hr = 40.28m/s
Speeder is moving at constant speed, then, a=0m/s²
After 2 seconds, the police car accelerate to 2.3m/s²
a = 2.3m/s²
Let assumed that the police car meets the speeder at d
Note, at the first 2 seconds,
The speeder will be at
d = ut + ½at², a=0
d = 40.28×2
d = 80.56m
And the police car will be at
d =ut = 25×2
d = 50m
So the speeder is already ahead at a distance of 80.56—50= 30.56m
So, let analyse from this new point and add the 2sec later,
Let assume the speeder traveled 'd' distance before they meet
Then, the police car will have to travel d+30.56 distance
Then, let calculate the time to reach distance, because they will spent the same time to reached there
For the speeder
d = ut
Then, d = 40.28t
For the police
d+30.56 = ut + ½at²
d = ut+½at² —30.56
d = 25t + ½•2.3•t²—30.56
d = 25t + 1.15t² - 30.56
So equating the two distances.
25t + 1.15t² - 30.56 = 40.28t
1.15t² —15.28t - 30.56 = 0
Divide through by 1.15
t² —13.29t—26.57 = 0
Using formula method to solve the quadratic equation
t = [-b±√(b²-4ac)]/2a
Where a = 1, b = -13.29 and c=-26.57
Solving this and discarding the negative time, we have
t = 15.05 s
Then, the total time the police will use to get to the speedy car is
t(total) = 15.05+2
t = 17.05 seconds