Question

You are driving at a constant speed on a straight highway that passes under an overpass. At time t=0 , when your car is a distance d from the overpass, a prankster on the overpass releases a water balloon from rest from a height h above the highway below.

Answer 1

Answer the following questions. You should treat the car as a point particle, ignoring its height, length, and width.

(a) Suppose that you don’t see the water balloon until it explodes on your windshield. At what speed v0 were you originally traveling, given that the water balloon strikes your car? Express your answer in terms of d, h, and any relevant constants. Simplify your answer as much as possible.

Answer 2

(b) Now suppose that you see the prankster release the water balloon at t=0 , and you slam on your brakes at time Tb in an attempt to avoid being hit. (Your reaction time is Tb>0 .) Your car slows down with a constant acceleration of magnitude As . The balloon hits the highway right in front of your car at the instant your car comes to rest. At what speed Vo were you originally traveling, given that the water balloon just barely misses your car? Express your answer in terms of d,h As,Tb and any relevant constants. Simplify your answer as much as possible.

Answer 3

(c) Let Va be the speed you found in part (a), and let Vb be the speed you found in part (b). Which of these two speeds is larger? Justify your answer mathematically, with comments as necessary.

Answer 4

@Callisto

Answer 5

Hello Callisto. I need help with this problem if you don't mind giving me a helping hand.

Answer 6

If you need I can explain what I did and you could probably guide me through it and tell me if i did anything wrong.

Answer 7

@hy123
Let's start with (a) - show me your attempt

Answer 8

@LastDayWork I used the formula x-x0 = (v0)t+1/2at^2 but the v0 = 0. So h=Vnaugh t +1/2 gt^2 therefore t = sqrt(2h/g)= Time the balloon takes to fall.

Answer 9

So far so good..

Answer 10

Since the car has to travel a distance, d, during the time, t, the ballon takes to fall, the time the car has traveled is d/v0 so i plugged that into h=1/2gt^2. From that i got v0= d sqrt(g/2h)

Answer 11

Is part (a) correct?

Answer 12

yep..now solve (b)

Answer 13

So part (a) is correct so far?

Answer 14

part (a) is correct..

Answer 15

For part (b) I let t=sqrt(2h/g) be the time the balloon takes to fall.

terms:
As= deceleration of the car
Tb= the reaction time.

d=(d-Vnaught (t-Tb)) = distance the car moves from the overall distance minus initial velocity times the time it takes to brake.

The final velocity of the car (V) is zero.

so I said d=Vnaught sqrt(2h/g)+1/2 (As) 2h/g

Answer 16

And to get the v0 the car was originally traveling I just get v0 on the left side, correct?

Answer 17

Are you getting something like this -
V(sqrt(2h/g) + (Tb)) = d + (As)h/g

Answer 18

What did you do with the 1/2?

Answer 19

It got cancelled as there was '2' in the numerator..

Answer 20

Ahh didn't catch it.

Answer 21

Lemme re-work it and see if i get that

Answer 22

Do you subtract Tb from the left side and add it to the right side and then divide both sides by sqrt(2h/g) to get v0 for that equation?

Answer 23

Can you write this equation again -
"...d=(d-Vnaught (t-Tb)) = distance the car moves from the overall distance minus initial velocity times the time it takes to brake...."

Answer 24

d = (d v0 (tb))

Answer 25

No, I mean the complete equation..mine was
d - V(Tb) = Vt - (1/2)At^2
where t = sqrt(2h/g)

Answer 26

Aren't we on part b?

Answer 27

t = sqrt(2h/g) is the time it takes the water balloon to fall.

Answer 28

Yep..the time it takes for the balloon to fell wont change by applying brakes XD

Answer 29

The distance the car moves is the overall distance minus the velocity initial times the braking time or reaction time.

so d=(d-Vnaught Tb)

Answer 30

I'm getting confused on this part then haha.