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 ?

Yaj143 Jan 26, 2021

#1**+1 **

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

CPhill Jan 26, 2021

#2**+1 **

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)

Guest Jan 27, 2021

#3**+1 **

...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

Guest Jan 27, 2021