Physics problem due Tomorrow!! please help me?

Question

A parachutist descending at a speed of 9.40m/s loses a shoe at an altitude of 31.60m.

(a) What is the velocity of the shoe just before it hits the ground?

Answer 1

OK, I come to your rescue.
So the initial velocity (we may take the x-direction downwards) of the shoe is v0 = 9.40m/s.
a = 9.8m/s^2 is the gravitational acceleration.
Let t be the time it takes the shoe to reach the ground. We have:
v0*t + a*t^2/2 = 9.40*t + 4.9*t^2 = 31.6
or: 4.9*t^2 + 9.40*t - 31.6 = 0
This is a quadratic equation, and the solutions can be found with formula x = (-b +- sqrt(b^2 - 4ac))/2 and they are:
t = -3.67 and t = 1.76 (s)
Ignore the negative solution and thus the velocity of the shoe just before it hits the ground is:
vo + a*t = 9.40 + 9.8*1.76 = 26.6 (m/s)

Answer 2

a million) a hundred and fifty=a million/2 x 9.8 x t^2 t=5.fifty 3 S=a hundred and fifty+a million.2 x5.fifty 3 s=156.6 m 2) Ball from suitable 20 = a million/2 x 9.8 x t^2 t=2.02 backside ball 20= u x 2.02 - a million/2 x 9.8 x 2.02^2 u=17.8 3) 20 x cos 30 = horizontal velocity 20/ (20 x cos 30) = 2/rt(3) seconds S=-25+20xsin30 x 2/rt(3) + a million/2 x 9.8 x 2^2/rt(3)^2 S=-25 + 20/rt(3) + 4.9 x 4/3

Answer 3

dv/dt = a = g
v2 = gx + v1
v2 = 31.6(9.8) + 9.40 = 319.08 m/s