A car of weight 1×104 N car comes to a bridge during a storm and finds the bridge washed out. The driver of weight 650 N must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.9 m above the river, while the opposite side is a mere 1.90 m above the river. The river itself is a raging torrent 52.0 m wide.
How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
Take the free fall acceleration to be g = 9.80 m/s^2.
What is the speed of the car just before it lands safely on the other side?
Take the free fall acceleration to be g = 9.80 m/s^2.
2006-06-28 18:27:06 · 7 answers · asked by GUVNA? 1 in Science & Mathematics ➔ Physics
double_nubbins has the right approach. I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2). Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.
Speed of the car? The horizontal speed? The falling speed? The speed in the direction of motion?
Horizontal speed: as calculated above
Falling speed = g*t, where t is the time calculated above.
Speed in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
2006-06-28 19:29:52 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
how long is the car?
one must land with most of it on the other side
what is the distance between the wheels?
the frount wheels would be falling while the rear wheels would still be on the ground. this would cause the car to flip end over end
what is the grade of the staring road?
drag on the car? how hard is it raining ?is there any wind ?what diretion is the wind
my guess is the car would never make it due to the fact that if the brigde slopes down hill 19 m over 52m the road on the high side would do the same
2006-06-28 18:47:42 · answer #2 · answered by specal k 5 · 0⤊ 0⤋
Projectile motion problem, I'd have to diagram it to explain it even decently, but suffice it to know that the car would have to be moving at 26.4 m/s or about 59 mph if he wants to survive. The car will hit the other side at 19.3 m/s or about 43.2 mph straight down, which might actually break something in his car anyway.
2006-06-29 08:27:06 · answer #3 · answered by The Frontrunner 5 · 0⤊ 0⤋
do now no longer use 'x' as a variable until finally it rather is given as such indoors the concern. to deal with the instruments, use them as multipliers of your areas, e.g. rather of 340 meters/2nd, write (340 x meters)/(a million x 2nd). This makes your concern as follows: fee = distance / time (340 x meters) / (a million x seconds) = distance / (3.5 x seconds) Now multiply the two aspects by potential of using the term 3.5 x seconds, so which you may comprehend a thank you to do from fractions in junior severe college, and you will see that seconds cancel out, leaving you a distance in meters, that's what you want. it rather is all indoors the 1st financial harm of your physics e book.
2016-10-31 21:46:02 · answer #4 · answered by ? 4 · 0⤊ 0⤋
ignore the weight.
He must travel 52 meters before he falls the difference in heights.
So 20.9-1.9= 19meters.
That makes it easy.
He must go 52 meters before he falls 19 meters.
Now you can figure it out yourself.
2006-06-28 18:33:35 · answer #5 · answered by double_nubbins 5 · 0⤊ 0⤋
Anyone who can solve this problem will get at least 10 thumbs up from me
2006-06-28 18:52:51 · answer #6 · answered by Sagely 4 · 0⤊ 0⤋
Didn't they try this in the movie "Roadtrip"?
2006-06-28 19:20:01 · answer #7 · answered by PC_Load_Letter 4 · 0⤊ 0⤋