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Calculating horizontal distance using angle and max height

  • Thread starter kgianqu2
  • Start date
  • #1
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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizontal, some of the tiny critters have reached a maximum height of 58.7 above the level ground.

A)What was the takeoff speed for such a leap?
I got 4.00 m/s which was correct.


B)What horizontal distance did the froghopper cover for this world-record leap?
I really am not even sure how to attempt this.
 

Answers and Replies

  • #2
326
3
kgianqu2 said:
A)What was the takeoff speed for such a leap?
I got 4.00 m/s which was correct.
Close! The answer should be 40 m/s. Use this form v_y = v_0 * sin(θ). Then, find d from this form:

v_f² = v_y² + 2ad

kgianqu2 said:
B)What horizontal distance did the froghopper cover for this world-record leap?
I really am not even sure how to attempt this.
Then, you will need to use this form:

v_x = v_0 * cos(θ)

This gives you the time. Now, find the time using this form: v_f = v_y + at [You need to use part (a) to answer this question!] Finally, using the value of v_x and the time you found (given v_y), find the horizontal distance traveled.

s = v_x * t
 
  • #3
16
0
When I answered 4.00m/s, mastering physics said it was correct.

I am not sure how to get time, but is v_x=36.7?
 
Last edited:

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