PHYSICS HELPPPPPPPPPPPPPP?

Question

1. A stone is dropped from a height of 50m. 2sec later a similar stone is throw down from the same height. If the two stone hit ground same time, wat was the approximate initial velocity of the second stone? ans: -36 m/s. ( My answer is -15.84 m/s, i can't get the answer given)

Answer 1

If g=9.81 m/s^2 then by using H=(1/2)g T^2 you have:

50=(1/2)(9.81)T^2

T^2=50/(1/2)(9.81)=100/9.81

T=3.19 sec.

For the secound stone you have H=(1/2)g t^2+V0 t formula.

in this formula t=T-2 or we have T=t+2

H=(1/2)g(T^2)=(1/2)g(t^2)+V0 t

(1/2)g(t+2)^2=(1/2)g(t^2)+V0 t

(1/2)gt^2+2gt+2g=(1/2)g^2+V0 t

2g(t+2)=V0 t

V0=2g(t+1)/t

t=T-2=3.19-2=1.19 sec.

V0=2(9,81)(2.19)/(1.19)=36 m/sec.

Answer 2

s=ut + (at^2)/2

s, displacement; u, inital velocity; t, time; a, acceleration

use the formula and simultaneous equation

s (first stone) = s (second stone)

ut + (at^2)/2 = u(t-2) + [a(t-2)^2]/2

s = ut + att/2   ,u=0

sqrt 50*2/9.8 = t

2 't's : t, t-2

2 u's : 0, u

a = 9.8 m/s/s

(9.8t^2) = u(t-2) + 9.8(t-2)^2/2

continue ... u have the t