A snow blower travels forward at a constant speed v_s = 1-4 ft/sec along the str
ID: 1717486 • Letter: A
Question
A snow blower travels forward at a constant speed v_s = 1-4 ft/sec along the straight and level path shown. The snow is ejected with a speed v_r = 30 ft/sec relative to the machine at the 40 degree angle indicated. Determine the distance d, which locates the snow-blower position from which an ejected snow particle ultimately strikes the slender pole P. At what height h above the ground does the snow strike the pole? Points O and Q are at ground level. Determine the location h of the spot toward which the pitcher must throw if the ball is to hit the catchExplanation / Answer
>> Now, Vs = Velocity of snow blower = 1.4 ft/sec = 1.4 i
>> Vr = Velocity of snow ejeted = 30 ft/sec at 40 degree, as shown
=> Vr = 30*sin40 j + 30*cos40 k = 19.28 j + 22.98 k
>> As, throw is made with respect to snow blower,
=> Vrs = Velocity of snow w.r.t snow blower = V = Vr - Vs = - 1.4 i + 19.28 j + 22.98 k
>> In x and z direction, there is no external force, So,
Acceleration, Ax = Az = 0
and, Ay = - 32.2 ft/s2
>> Along X - Direction,
distance travelled = d , let in time = t
=> d = Vx*t = -1.4*t .......... (1)....
>> Along Y - Direction,
distance travelled = h - 3 feet, let in time = t [ h is in ft ]
>> According to newtons's 2nd Equation of motion,
s = ut + (1/2)*a*t2
=> h - 3 = 19.28*t - 0.5*9.81*t2 ...... (2)....
>> Along Z - Direction,
distance travelled = 15 - 1.5 feet = 13.5 ft , let in time = t
=> 13.5 = Vz*t = 22.98*t
=> t = 0.588 sec
From (1).
d = - 1.4*0.588 = 0.82 ft
-ve represents that Q pint lies beyond O point at a distance, d = 0.82 ft
>> From (2).
h - 3 = 19.28*t - 0.5*32.2*t2 [ time, t = 0.588sec ]
=> h = 8.77 ft .......ANSWER .........
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