Ironman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/ 3 in the last interval of time of 1.1 s of his fall.
Hint: First, compute the velocity when Ironman reaches the height equal to the distance fallen. This requires that you do the following: define origin as the bottom of the building. Then use x-x0 = -v0*(t-t0)-(1/2)g(t-t0)^2 where x=0 and x0= (distance fallen) and t-t0 is the time interval given. In this formulation, you are going to get magnitude of v0 since you already inserted the sign.
You then insert v0 that you just calculated into the kinematic equation that involves v, g, and displacement (v^2-v0^2 = 2g(height-(distance fallen)), but now v (which is the final velocity is v0 from above) and v0 in this case is the velocity that the Ironman has when he begins to fall, which is 0.
This gives a quadratic equation for height h, and you will need to use the binomial equation to solve for h. Choose the larger of the two solutions.
Theres a simpler way to do this, since there is zero initial velocity. We might as well use the frame of reference , the top of the building is our origin. And let down movement be positive.
Since there is zero initial velocity. We might as well use the top of the building as our origin. And let down movement be positive , so he accelerating in the same direction as his motion.
The equation for the distance the ironman moved
is y = 1/2 gt^2
Let h = total height, and T be total time.
h = 1/2 gT^2
Now it tells us that when T - 1.1 sec the ironman is h/3 units of distance from the ground. This is the same thing as saying he covered h*2/3 in T-1.1 seconds.
2/3 *h = 1/2 g ( T - 1.1)^2
now its simpler if we substitute 1/2gT^2 for h. The other way is more tedious because of the square rooting.
2/3 * (1/2 gT^2) = 1/2 g ( T - 1.1)^2
2/3 T^2 = (T - 1.1)^2
2/3 T^2= T^2 - 2T*1.1 + 1.1^2
0 = 1/3 T^2 - 2.2T + 1.21
0 = T^2 -6.6 T + 3.63
This gives us two solutions T= 5.9944 ,0.6056 approximately
The second solution we reject because we want T - 1.1 , and that will make it negative.
So that's T, and using 9.8 meters /s^2 for g we have
h = 1/2 9.8 (5.9944)^2 meters
and height is roughly 176 meters high to the nearest meter
Why did Ironman fall from the building? I guess he was just "a-fall-ed" to do it!
To solve this problem, we need to follow the given hints and steps. Let's start step by step:
Step 1: Define the origin as the bottom of the building and let x be the distance fallen.
Step 2: Use the equation x - x0 = -v0 * (t - t0) - (1/2)g(t - t0)^2, where x = 0, x0 = distance fallen, t - t0 = 1.1 s, and g = acceleration due to gravity (approximately 9.8 m/s^2).
This gives us the equation 0 - x0 = -v0 * (1.1) - (1/2) * 9.8 * (1.1)^2.
Step 3: Solve the equation for v0. Rearranging the equation, we get:
v0 = (x0 / 1.1) - (1/2) * 9.8 * (1.1)
Step 4: Insert v0 into the kinematic equation v^2 - v0^2 = 2g(height - distance fallen), where v0 = 0 (since Ironman starts from rest).
This gives us the equation 0 - v0^2 = 2 * 9.8 * (height - distance fallen).
Step 5: Substitute v0 from Step 3 into the equation. We get:
0 - [(x0 / 1.1) - (1/2) * 9.8 * (1.1)]^2 = 2 * 9.8 * (height - distance fallen).
Simplifying the equation, we have:
21.56 - (x0 / 1.1) + 9.8 * (1.1)^2 = 19.6 * height
Step 6: Rearrange the equation and solve for height. We have:
height = (21.56 - (x0 / 1.1) + 9.8 * (1.1)^2) / 19.6
Step 7: Evaluate the expression using the given value of x0 (h/3) and calculate the height.
height = (21.56 - ((h/3) / 1.1) + 9.8 * (1.1)^2) / 19.6
Step 8: Solve the equation for height. You will need to use the binomial equation to find the roots of the quadratic equation.
The value of h can be determined by solving the quadratic equation obtained in Step 7. Using the binomial equation, we get:
h = [-b ± √(b^2 - 4ac)] / (2a)
where a = 19.6, b = -(1/3) / 1.1, and c = 21.56 + 9.8 * (1.1)^2.
Step 9: Substitute the values into the equation and calculate the roots.
h = [((1/3) / 1.1) ± √(((1/3) / 1.1)^2 - 4(19.6)(21.56 + 9.8 * (1.1)^2))] / (2 * 19.6)
Step 10: Calculate the values of h using the quadratic formula.
Once you plug in the values into the quadratic formula, you will obtain two possible values for h. Choose the larger of the two solutions.
Note: The exact numerical calculations will depend on the specific values given for x0 and g.
To solve this problem, we'll follow the steps provided in the hint:
Step 1: Define the origin and establish a coordinate system. Let's assume the bottom of the building is the origin, with the upward direction positive.
Step 2: Use the equation x - x0 = -v0 * (t - t0) - (1/2) * g * (t - t0)^2, where x = 0 since Ironman reaches the ground, x0 = h (the distance fallen), and t - t0 = 1.1 s. We want to solve for v0.
Since x = 0 and x0 = h, the equation becomes h = -v0 * (1.1 s) - (1/2) * g * (1.1 s)^2.
Step 3: Rearrange the equation to solve for v0: h = -1.1 * v0 - (1/2) * g * (1.21 s^2).
Step 4: Now, we can use the kinematic equation: v^2 - v0^2 = 2g * (height - distance fallen), where v = 0 (final velocity when Ironman reaches the ground) and v0 is the velocity when Ironman begins to fall (which we solved for in Step 3). Note that height = h and distance fallen = h/3.
Plugging in the values, we get 0 - v0^2 = 2g * (h - h/3).
Simplifying, -v0^2 = 2g * (2h/3).
Step 5: Substitute the calculated value of v0 from Step 3 into the equation to get -(-1.1 * v0 - (1/2) * g * (1.21 s^2))^2 = 2g * (2h/3).
Step 6: Solve the quadratic equation for h using the binomial equation or any other appropriate method. Choose the larger of the two solutions.
By following these steps, you will be able to calculate the distance fallen (h) by Ironman.