A Balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of descending balloon and 4.00 (s) later this person sees the stone strike the ground.

Question

A) How high was the balloon when the stone was thrown out?
B) How far horizontally does the stone travel before it hits the ground?
C) At the instant the stone hits the ground, how far is it from the basket?

Answer 1

u = 15 forever

Vi = -20

h = Hi + Vi t + (1/2) a t^2
0 = Hi - 20 t - 4.9 t^2
but t = 4
so solve for Hi, initial height

in 4 seconds at 15 goes 60 meters

where is the basket at 4 seconds?

h basket = Hi - 80

d = sqrt (60^2 + h basket^2)

Answer 2

is your d = square root of (60^2 + h basket^2) and for which question a,b or c