Question on TBR 1.15

[l0lamd]

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The fruit juggler, dissatisfied with simply dropping fruit, decides to throw an apricot off a cliff. The apricot leaves the top of the cliff at a speed of 35 m/s and at an angle of 50 degrees above the horizontal. If the cliff is 310m tall, how far down range is the fruit when it hits the ground?

Is there a reason I can't use the r=[2(initial velocity in x direction)(initial velocity in y direction)]/g equation?

 

[Arctangent]

The short answer is that the equation assumes the elevation difference between your throwing point and your ending point are the same. In other words, you cannot use this equation if you are starting your throw from a cliff or a trough relative to the end point. This question states that the start point is 310m above, so the equation won't work.

The derivation is:
t_flight = (v_final - v_initial)/g .... The velocities here are in the vertical direction.
v_final = -v_initial .... This step is the one that assumes there is no cliff or trough
t_flight = (v_initial + v_initial)/g = 2(initial velocity in y direction)/g
range = t_flight * initial velocity in x direction
range = 2(initial velocity in x direction)(initial velocity in y direction)/g

 

[pm1]

The short answer is that the equation assumes the elevation difference between your throwing point and your ending point are the same. In other words, you cannot use this equation if you are starting your throw from a cliff or a trough relative to the end point. This question states that the start point is 310m above, so the equation won't work.

The derivation is:
t_flight = (v_final - v_initial)/g .... The velocities here are in the vertical direction.
v_final = -v_initial .... This step is the one that assumes there is no cliff or trough
t_flight = (v_initial + v_initial)/g = 2(initial velocity in y direction)/g
range = t_flight * initial velocity in x direction
range = 2(initial velocity in x direction)(initial velocity in y direction)/g

I saw that in the book TBR solution solves through a quadric equation after going through y = vot+at^2/2

Is there an alternative not using quadratic equation?

I was trying to solve by doing the range as if it would come back to the same level that it left of, and then add an extra representing the range for the 310m below the cliff but then I wasn't sure on what to put Voy as.

 

[Arctangent]

As for an alternative, you could definitely do it the way you suggested, and avoid the quadratic formula. You use the formula for the first part of the flight. For the rest of the flight path, you have:

Initial vertical velocity = initial vertical velocity at beginning, just opposite direction.

You can first get final velocity at impact using: (final vertical velocity)^2 = (initial vertical velocity)^2 + 2gx, where x is the height, 310m.

You can get time duration of the second part via: time = (final vertical velocity - initial vertical velocity)/g

And for range of the second part, use: r = (horizontal velocity)*time

If you do use the quadratic formula and y = vot+at^2/2, you can combine those first two steps into one.