smc-1159-60-Terminal velocity

A helicopter is hovering at a constant height above a fixed location. A skydiver falls from rest for two seconds from the helicopter. The skydiver is subject only to gravitational acceleration and air resistance is negligible for the first two seconds. Let downward displacement be positive.

Find the distance, in metres, fallen in the first two seconds. (2 marks)
Show that the speed of the skydiver after two seconds is 19.6 ms^–1. (1 mark)

After two seconds, air resistance is significant and the acceleration of the skydiver is given by `a = g -0.01v^2`.

Find the limiting (terminal) velocity, in ms^–1, that the skydiver would reach. (1 mark)
i. Write down an expression involving a definite integral that gives the time taken for the skydiver to reach a speed of 30 ms^–1. (2 marks)
ii. Hence, find the time, in seconds, taken to reach a speed of 30 ms^–1, correct to the nearest tenth of a second. (1 mark)
Write down an expression involving a definite integral that gives the distance through which the skydiver falls to reach a speed of 30 ms^–1. Find this distance, giving your answer in metres, correct to the nearest metre. (3 marks)

Show Answers Only

`19.6\ text(m)`
`text(See Worked Solutions)`
`v_t = 10sqrtg\ text(ms)^(−1)\ \ text(downwards)`
i. `t(30) = int_19.6^30 100/(100g – v^2)dv + 2`
ii. `5.8\ text(s)`
`120\ text(m)`

Show Worked Solution

a. `u = 0, \ t = 2, \ a = g = 9.8`

`x`	`= ut + 1/2 at^2`
	`=0 xx 2 + 1/2 xx 9.8 xx 2^2`
	`= 19.6\ text(m)`

b.	`v`	`= u+ at`
	`v(2)`	`= 9.8 xx 2`
		`= 19.6\ text(ms)^(−1)\ \ \ text{(downward → positive)}`

c. `text(Terminal velocity), v_t, text(occurs when)\ \ a = 0,`

♦ Mean mark 48%.

`g -0.01v_t^2`	`=0`
`v_t^2`	`= 100 xx 9.8`
`:. v_t`	`= 14 sqrt5\ text(ms)^(−1)\ \ \ (v_t > 0\ \ text{as}\ \ v_t\ \ text{is downwards})`

d.i.	`(dv)/(dt)`	`= g – 0.01v^2`
	`(dt)/(dv)`	`= 1/(g – 0.01v^2)`
		`= 100/(100g – v^2)`
	`t`	`= int 100/(100g – v^2)`

`text(Time taken until)\ \ v=19.6\ \ text(is 2 seconds.)`

♦ Mean mark part (d)(i) 39%.

`:.\ text(Time taken until)\ \ v=30`

`=int_19.6^30 100/(100g – v^2)\ dv + 2`

d.ii. `5.8\ text(seconds)\ \ \ text{(by CAS)}`

♦♦ Mean mark part (d)(ii) 25%.

e.	`v *(dv)/(dx)`	`= g – 0.01 v^2`
	`(dv)/(dx)`	`= g/v – (v^2)/(100v)`
		`= (100g – v^2)/(100v)`
	`(dx)/(dv)`	`= (100v)/(100g – v^2)`
	`x`	`= int (100v)/(100g – v^2)\ dv`

`text(Distance fallen in 1st 2 seconds)\ = 19.6\ text(m)`

♦♦ Mean mark part (e) 29%.

`:.\ text(Total distance fallen until)\ \ v=30`

`= int_19.6^30 (100v)/(100g – v^2)\ dv + 19.6`

`~~ 120\ text(m)`