A) How fast should the car be traveling just as it leaves the cliff in order jus

Question

A) 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?

B) What is the speed of the car just before it lands safely on the other side?

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 18.7 m above the river, while the opposite side is a mere 1.3 m above the river. The river itself is a raging torrent 69.0 m wide

Explanation / Answer

1)

along horizontal
________________

initial velocity v0x = v

acceleration ax = 0

initial position = x0 = 0

final position = x = 69

displacement = x - x0 = 69 m

from equation of motion

x - x0 = v0x*T + 0.5*ax*T^2

x - x0 = v*T

T = (x - x0)/v......(1)

along vertical
______________

v0y = 0

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 18.7 m

final position y = 1.3 m

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = - (0.5*g*(x-x0)^2)/v^2)

1.3 - 18.7 = -0.5*9.8*69^2/v^2

speed v = 36.6 m/s <<<-------------ANSWER

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part B

along horizontal

vx = vox + a*t = v = 36.6 m/s

along vertical

vy^2 - voy^2 = 2*ay*(y-y0)

vy^2 - 0 = -2*9.8*(1.3-18.7)

vy = 18.5 m/s

speed of the car before landing v = sqrt(v^2+vy^2)

v = sqrt(36.6^2+18.5^2)

v = 41 m/s <<<-------------ANSWER

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