Calculating horizontal distance using angle and max height

AI Thread Summary
The discussion focuses on calculating the horizontal distance a froghopper can jump using its takeoff speed and angle. The correct takeoff speed for the jump is 4.00 m/s, confirmed by one participant. To determine the horizontal distance, participants are guided to use the equations for vertical and horizontal motion, specifically involving the sine and cosine of the launch angle. There is some confusion regarding the calculation of time and horizontal velocity, with one participant suggesting a horizontal velocity of 36.7 m/s. The thread emphasizes the importance of using the correct formulas to derive the necessary values for calculating the jump's horizontal distance.
kgianqu2
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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.
 
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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
 
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?
 
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