Please help with this question

Question

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 48.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.

(a)

the frog's launch speed (in m/s)

  m/s

(b)

the time the frog spends in the air (in s)

  s

Answer 1

Ok, let's take frog's launch speed V_launch.

It has two compoents as the vertical (subject to gravitational acceleration) and horizontal (constant neglecting air resistance). Vertically, it is V_y = V_launch x sin48.5 and horizontal V_x = V_launch x cos48.5

The time of travel t can be found by dividing the 2.20/Vx;

So, t = 2.20/V_x.

a) Vertically, the equation V_final = V_initial + at, let's substitite the values below.

-V_launch x sin48.5 = +V_launch x sin48.5 - (9.8 x 2.20/V_launch x cos48.5)

21.56/( V_launch x 0.66) = 2 x V_launch x 0.75

32.7/ V_launch = 1.5 V_launch

21.8 = V_launch²

V_launch = 4.67 m/s

b) t = 2.20/Vx = 2.20/V_launch x cos48.5

t = 0.71s