Arturo O. answered • 10/04/17
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I assume the 60 m/s is horizontal speed. Find how long it takes for the bomb to travel a horizontal distance of 900 m at constant horizontal speed u = 60 m/s, plug that time in the height equation with the height set to zero, and solve for the initial vertical speed v.
h(t) = -4.9t2 + vt + 1100
The target is reached when
t = (900 m)(60 m/s) = 15 s
0 = h(15) = -4.9(15)2 + 15v + 1100 ≅ 15v - 2.5
v = 2.5/15 m/s ≅ 0.1667 m/s (in the upward direction)
Test the answer. Is ut = 900 when t = 15 s?
ut = 60(15) = 900 [good]
Is h(t) = 0 when t = 15 s ?
h(15) = -4.9(15)2 + 0.1667(15) + 1100 ≅ 0 [good]
Note that the bomb must be launched up rather than released down.
Arturo O.
Under the conditions of this problem, the solution gives an initial upward vertical velocity. Note that h(t) with an initial upward speed puts the bomb on the target's y-coordinate at the same time that x(t) puts it on the target's x-coordinate.
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10/04/17
Gnarls B.
10/04/17