Question

Spiderman while diving with a velocity of 20 m/s from the top of a 100 m building notices that he has run out of "Spider Web". To cushion his fall he grabs on to a horizontal flag pole which bent when he grabs on to it.

(a) If his velocity the instant he grabbed on to the flag pole was 40 m/s. Determine how far below the top of the building the flag pole was located.

(b) If the flag pole bent and cushioned his fall such that it slowed him down to 1/4 of his original velocity and then broke. Determine his total time of fall from the instant he jumped off the building to when he hit the ground.​

Answer 1

Let's solve each part of the problem separately.

**(a) How far below the top of the building the flag pole was located:**

We can use the equations of motion for uniformly accelerated motion to find the displacement. The equation is given by:

\[ v_f^2 = v_i^2 + 2a d \]

Where:
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity,
- \( a \) is the acceleration,
- \( d \) is the displacement.

Given that Spiderman's initial velocity (\(v_i\)) is 20 m/s, his final velocity (\(v_f\)) is 40 m/s, and the acceleration (\(a\)) is due to gravity (approximately -9.8 m/s²), we can rearrange the equation to solve for \(d\):

\[ d = \frac{v_f^2 - v_i^2}{2a} \]

Substitute the values and solve for \(d\).

**(b) Total time of fall from the instant he jumped off the building to when he hit the ground:**

The total time of fall can be found by adding the time taken to reach the maximum height and the time taken to fall from that maximum height to the ground.

1. **Time to reach maximum height:**
Use the equation:
\[ t_{\text{max}} = \frac{v_f - v_i}{a} \]

2. **Time to fall from maximum height to the ground:**
The time of free fall can be found using:
\[ t_{\text{fall}} = \sqrt{\frac{2h}{g}} \]
where \( h \) is the height (100 m) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

The total time of fall is \( t_{\text{max}} + t_{\text{fall}} \).

Answer 2

(a) Approximately
`61.2\; {\rm m}` from the top of the building.

(b) Approximately
`4.0\; {\rm s}`.

Assumption:
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Approach this question in the following steps:

• Find displacement during the first part of the motion.
• Find the additional distance that needs to be covered in the second part of the motion, which is equal to the height of the flagpole above the ground.
• Find the duration of each part of the motion.

(a)

Between starting the dive and grabbing the horizontal flag pole, vertical acceleration is constantly
`a = g = 9.81\; {\rm m\cdot s^(-2)}`. During this part of the motion:

• Initial velocity is
  `u = 20\; {\rm m\cdot s^(-1)}`,
• Final velocity is
  `v = 40\; {\rm m\cdot s^(-1)}`, and
• Displacement
  `x` needs to be found.

Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find displacement. Rearrange this equation to obtain:

`\begin{aligned}x &= (v^(2) - u^(2))/(2\, a) \\ &= ((40)^(2) - (20)^(2))/(2\, (9.81))\; {\rm m} \\ &\approx 61.2\; {\rm m}\end{aligned}`.

Hence, the distance between the top of the building and the horizontal flag pole is approximately
`61.2\; {\rm m}`.

(b)

It is given that the height of the building is
`100\; {\rm m}`. Since the distance between the top of the building and the horizontal flagpole is approximately
`61.2\; {\rm m}`, the flagpole would be approximately
`(100 - 61.2)\; {\rm m}\approx 38.8\; {\rm m}` above the ground.

Between breaking the flagpole and landing on the ground:

• Initial velocity is
  `u = 10\; {\rm m\cdot s^(-1)}`, and
• Displacement is
  `x \approx 38.8\; {\rm m}`.

The duration
`t` of this motion can be found in the following steps:

• Find final velocity
  `v`.
• Divide the change in velocity
  `(v - u)` by the acceleration to find the duration
  `t` of the motion.

Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find final velocity:

`\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \\ &\approx \sqrt{(10)^(2) + 2\, (9.81)\, (38.8)}\; {\rm m\cdot s^(-1)} \\ &\approx 29.360\; {\rm m\cdot s^(-1)}\end{aligned}`.

Divide the change in velocity by acceleration to find the duration of the motion:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx (29.360 - 10)/(9.81)\; {\rm s} \\ &\approx 1.97\; {\rm s}\end{aligned}`.

Similarly, the duration of the first part of the motion would be:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx (40- 20)/(9.81)\; {\rm s} \\ &\approx 2.04\; {\rm s}\end{aligned}`.

Hence, the total duration of the motion would be:

`(1.97)\; {\rm s} + (2.04)\; {\rm s} \approx 4.0\; {\rm s}`.