Exercise 3.13 C14 of Part A A car comes to a bridge during a shorm and finds the

Question

Exercise 3.13 C14 of Part A A car comes to a bridge during a shorm 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 18.4 m above the river, whille the opposibe side is a mere 14 m above the river The river itsolf is a raging torrent 560 m wide How fast should the car be traveling just as it leaves the clif in order just to clear the fiver and land safoly on the opposie side? Part B What is the speed of the car just before it lands safely on the other side?

Explanation / Answer

Horizontal distance traveled by car = 56 m

vertical distance traveled by car = 18.4 - 1.4 = 17 m

ax = Horizontal acceleration = 0 m/sec^2

ay = vertical acc. = +*9.81 m/sec^2

V0x = V0 m/sec

V0y = 0 m/sec

Using the vertical motion

y = V0y*t + 0.5*g*t^2

17 = 0*t + 0.5*9.81*t^2

t = sqrt (2*17/9.81) = 1.86

horizontal distance traveled in this time

x = V0x*t + 0.5*ax*t^2

x = V0*1.86 + 0.5*0*1.86^2

V0 = x/t

V0 = 56/1.86= 30.10 m/sec

B.

Vx = 30.10 m/sec

Vy = V0y + ay*t

Vy = 0 + 9.81*1.86 = 18.25 m/sec

Total velocity will be

V = sqrt (Vx^2 + Vy^2)

V = sqrt (30.10^2 + 18.25^2) = 35.20 m/sec

I hope help you.....

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