+61 an arrow is shot upward on the moon, with a velocity of 70 m/s its height (in meters ) after t seconds is given by H(t) = 70t - 0.99t ^ 2 With what velocity will the arrow hit the moon ?
+129899 We need to find out when the height = 0
70t - .99t^2 = 0 factor
t (70 - .99t) = 0
The scond factor set to 0 will give us the time we need
70 - .99t = 0
70 = .99t
t = 70 /.99 (sec)
If you have had Calculus, the velocity function is
H'(t) = 70 - 1.98 t
The velocity at 70/.99 sec is
70 - 1.98 ( 70/.99) =
= - 70 m / s
With no air friction, the velocity magnitude up will = the velocity magnitude down
70 m/s up initially will result in 70 m/s down (or -70 m/s)
...another way to look at it... the initial KE is all converted to Potential energy....then the arrow stops and begins to fall back down...
now all of the PE is coverted to KE (same as the origianal) ....so it will have the same magnituide velocity when it hits the ground as it had going up (though in opposite direction) -70 m/s
