Calculating horizontal distance using angle and max height

[kgianqu2]

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.

 

[NasuSama]

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

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

 

[kgianqu2]

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?