A snow blower travels forward at a constant speed v_s = 1-4 ft/sec along the str

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 catch

Explanation / 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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