a physic question>> i'm so confuse?

Question

A car manufacture tests its cars for front end collision by hauling them up on a crane and dropping them from a certain height.
a.) show what the speed just before a car hits the ground, after falling from rest a vertical distance H, is given by (Square root of 2gH) what height correspinds to a collision at (b.) 50km/h? (c.) 100 km/h

Answer 1

Two ways to do this. Let's use kinematics, then the relationship between distance, speed and acceleration is:

y = (v^2 - v0^2)/(2a) v0 = 0 in this problem and a = g, y = H

H = v^2/(2g) ---> v = sqrt(2gH)


You can do the arithmetic.

Answer 2

NO the two the balls have diverse very final velocities.... evaluate the third equation of action v * v (sq. of very final speed) = u * u (sq. of preliminary speed) + 2gh h: height lined on a similar time as falling right here hence -the ball that's dropped has u = 0 and is on the peak think "H" -for the 2nd ball , it first is going upwards to a optimal height enable "x" and then comes right down to the floor that's now on the peak of (H +x) from that component. now pondering the two the factors having preliminary speed 0 , then the third equation could be written as v * v (sq. of very final speed) = 2gh hence , v * v (sq. of very final speed) is immediately proportional to "h" that's the peak lined. using fact the peak ( H + x ) is obviously extra beneficial than the H hence the astounding speed for the ball which one is thrown upwards would be extra. desire you recognize WHAT i choose for to declare . SORRY FOR ANY MISTAKE

Answer 3

a) 50km/h= sqrt(2(9.8m/s^2)(H)
converting 9.8m/s^2 to km/h^2
9.8m/s^2 * 3600s/1hr *3600s/1hr * 1km/1000m=127008km/hr^2

50km/h=sqrt(127008km/hr)(H) square both sides

2500km^2/hr^2 = (127008km/hr^2)(H)
H = 0.019683km
H= 20m

since the height is doubled ang the velocity is directly proportional to the square of the height the H wil be quadrupled therefor for b) H=20*4=80m