If a stone is thrown vertically upward from the surface of the moon with a velocity of 5 m/s, its height (in meters) after t seconds is h = 5t - 0.83t^2.
(a) What is the velocity of the stone after 1 s?
v = m/s
(b) What is the velocity of the stone after it has risen 5 m?
v = (smaller value)
v = (larger value)
2007-10-20 03:40:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
a) v(t)= 5-1.66t
t=1, v(1)=3.34 m/s
b) When h=5, 5=5t-.83t^2, it becomes
.83t^2-5t+5=0
Use quadratic formula will give you t=1.27 s or t=4.758 s
v(1.27)=5-1.66(1.27)=2.89 m/s
v(4.758)=5-1.66(4.758)=-2.98 m/s which is on its way down.
2007-10-20 04:30:29 · answer #1 · answered by mlam18 6 · 0⤊ 0⤋
velocity is given by differiatiating h to t
so v = dh/dt = 5-1,66t
putting t=1 we get
v=5-1.66 = 3.34 m/s
(b) when h=5 in given eqn we get
5=5t-0.83t^2
or 0.83t^2-5t+5=0
Solve the equation for t using quadratic formula.
Then using above expression
v=5-1.66t, you get two values of v.
Do it yourself for practice.
2007-10-20 04:06:08 · answer #2 · answered by mwanahamisi 3 · 0⤊ 0⤋
a) v(t) = 5 - 1.66t; t=1 thus v = 3.34 m/s
b) IF it has risen 5 meter THEN the velocity will be smaller, but maybe it wont rise 5 meter at all.
2007-10-20 03:47:44 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
v=dh/dt = 5-1.66 t so after 1 s v= 3.34 m/s
t for maximum height = 5/1.66 = 3.01 s
0.83t^2 -5t+5 =0 so t = ((5+-sqrt(25-16.6))/1.66=1.27 s or4.76s
v up =2.89 m/s
v down = -2.89 m/s
2007-10-20 04:05:05 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋