A rescue plane wants to drop supplies to isolated mountain climbers on a rocky r

Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. Assume the plane is traveling horizontally with a speed km/h (77.8 m/s). (a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Figure a)? 538.8 m (b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be so that they arrive precisely at the climbers' position (Figure b)? (Assume up is positive.) m/s m/s On mountainous downhill roads, escape routes are sometimes placed at the side of the road for trucks whose brakes might fall. Assuming a constant upward slope of 32 degree, calculate the horizontal vertical components of the acceleration of a truck that slowed from 150 km/h to rest in 6.0 s. See the figure below. (Assume up and right are positive.)

Explanation / Answer

Vx= 77.8 m/s

Dx= 425 m

time= 425/ 77.8=5.463 seconds apprx

U y = initial vetival veolcity

225= u(5.463) + 0.5 (9.8) (5.463)^2

u =14.42 m/s

V = sqroot ( 14.42 ^2 +77.8^2) = 79.125 m/s

angle = tan^-1 ( 225/425) =27.9 degree apprx

initial vertical velocity = 79.125 sin 27.9 =37.02 m/s( below the horizontal)

final vertical velocity= 37.02 + 9.8 (5.463 ) =90.557 m/s

Final velocity ( at landing) = sqroot(77.8^2 + 90.557^2)= 119.39 m/s apprx

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