# 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
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:

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

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?

Last edited: