Question

An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 160 m above the level ground when the plane is flying at 75.0 m/s 55.0 ∘ above the horizontal. a) How far in front of the cattle should the pilot release the hay so that the bales will land at the point where the cattle are stranded? b)

Answer 1

The horizontal distance traveled by the bales before it reaches the cattle is 95.1 m

The given parameters;

• height reached by the bale, h = 160 m

• velocity of the bale, v = 75 m/s

• angle of projection, θ = 55.0 ⁰

The time of flight of the bales before it reaches the cattle is calculated as;

$h = v_o_yt + \frac{1}{2} gt^2\\\\160 = (75\times sin(55) )t \ + \ (0.5\times9.8)t^2\\\\160 = 61.425t + 4.9t^2\\\\4.9t^2 + 61.425t - 160 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ b = 61.425, \ c = -160\\\\t = \frac{-b \ + /- \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-61.425 \ + /- \ \sqrt{(61.425)^2 - 4(4.9\times -160)} }{2\times 4.9}\\\\t = 2.21 \ s$

The horizontal distance traveled by the bales before it reaches the cattle is calculated as;

$R = v_o_x t\\\\R = v\times cos(\theta) \times t\\\\R = 75 \times cos(55) \times 2.21\\\\R = 95.1 \ m$

Thus, the horizontal distance traveled by the bales before it reaches the cattle is 95.1 m

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