An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground...
An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?
1 Answer
Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator
Hello!
The height of the first object is given by
`h1(t) = 60 - 10*t - (g*t^2)/2` ,
where 60 is the initial height, -10 is the initial downward speed and g is gravity acceleration, g=`9.8m/s^2` .
The height of the second object is
`h2(t) = 40t`
(I suppose its speed is assumed to be constant).
Note that t>=0. Objects pass each other when they have equal height, i.e. h1(t) = h2(t):
`60 - 10*t - (9.8*t^2)/2 = 40t` , or
`4.9*t^2 + 50*t - 60 = 0` .
The roots are ` ``t1,2 = (-50 +- sqrt(2500 + 4.9*240))/9.8 approx (-50 +- 60.63)/9.8` .
The root with "-" is <0, the only solution is t1 = 10.63/9.8 = 1.085 (s).
The heigth at t1 is 40*t1 = 40*1.085 = 43.4 (m).