# Question 0a476

Jul 3, 2015

The minimum acceleration needed is ${\text{4.44 m/s}}^{2}$.

#### Explanation:

FULL QUESTION

From a motionless helicopter a parachute jumper jumps with a motionless position (which means that the velocity is ${\text{0 ms}}^{- 1}$).

He tries to open the parachute, but it does not open in the needed moment. There is a vehicle with a metros for his safety stopped 500m away from the point he is about to fall.

At this particular moment the vehicle's machines show that he is falling at a velocity of "10 ms^(-1) and is 600m up from the ground.

(Consider that the vehicle can stop immediately at any location it wishes to, the resistance of the air is zero and the gravitational acceleration is "10 ms^(-2)).

At the particular moment the vehicle starts moving. To come to the place where the parachute jumper is about to fall what has to be the minimum constant acceleration of the vehicle?

The idea behind this problem is that the ground vehicle must cover 500 m, the distance between its current location and where the paratrooper will land, in the same time it takes the paratrooper to cover 600 m in free fall.

At the point when the parachute fails to open, the paratrooper has a velocity of 10 m/s. This means that it will cover 600 meters in

$h = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{\text{=10 m/s}}} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$600 = 10 \cdot t + 5 \cdot {t}^{2}$

This is equivalent to

${t}^{2} + 5 t - 300 = 0$

This equation will produce two values for $t$, a positive one and a negative one. Pick the positive one to get

$t = \text{15 s}$

This means that the vehicle needs to cover 500 m in 15 seconds, which means that its acceleration will be

d = underbrace(v_0)_(color(blue)("=0)) * t + 1/2 * a * t^2#

${\text{500 m" = 1/2 * a * 15^2"s}}^{2}$

$a = \left(2 \cdot {\text{500 m")/(225"s"^2) = color(green)("4.44 m/s}}^{2}\right)$