A small object begins a free-fall from a height of H=85.0 m at t0=0 s . After =2

Question

A small object begins a free-fall from a height of H=85.0 m at t0=0 s . After =2.40 s , a second small object is launched vertically up from the ground with an initial velocity of v0=40.8 m/s . At what height from the ground will the two objects first meet?

r=

theta=

part b)

Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a =49.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s2 , find the displacement vector r and the angle of the displacement 1 .

Enter the components of the vector, rx and ry , and the angle. Assume the time interval is still 3.50 s .

rx=

ry=

theta=

r =rxi +ryj

Explanation / Answer

1) Y1(t) = 85 - 4.9t^2

Y2(t) = 40.8t - 4.9t^2

Y1(t)= Y2(t) ===> t = 2.08 s

Plugging in t=2.052 s into Y1(t) or Y2(t), we get the answer: 63.7 m

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