Question

object kicked at 45 degree and travels 82 m b4 hitting the ground. what was its vo? how long was it in the air? how high did it go?

please and thanks

Answer 1

Neglect air resistance, although it is certainly important for kicked objects.

The distance an object travels when it starts at velcoity Vo and angle A from the horizonal is (Vo^2/g) sin (2A) Set that distance equal to 82 m and solve for Vo.

The time in the air is
T = (82 m)/Vo cos A

The maximum height can be obtained from
(1/2) g (T/2)^2 = H,
since the object spends T/2 falling, starting with zero vertical velocity at the top of the trajectory.

Answer 2

Gee, you want to know it all!
well, at 45 degrees
initial vertical speed = .707 Vo
constant horizontal speed = .707 Vo as well
It goes 82 meters horizontal so
84 = .707 Vo T
T = 119/Vo
where T is total time aloft
Now the vertical problem
v = .707 Vo - 9.8 t
v = 0 at top
so
t = .0721 Vo at top
(We all know t = 1/2 T but we will not admit that we know that)
height = 0 + .707 Vo t - 4.9 t^2
height = .707 Vo (.0721 Vo) -4.9(.00520)Vo^2
height = (.0510-.0255)Vo^2
height = .0255 Vo^2
Now our last equation relates to T and the total vertical flight until we are back on the ground.
0 = 0 + .707 Vo T - 4.9 T^2
T = 0 is a solution, but that was on the ground when we started so we want
T = .144 Vo
but we already know that T = 119/Vo
so
Vo*.144 = 119/Vo
so
Vo = 28.7 m/s <---- Finally an answer
T = time in air = Vo*.144 = 4.14 seconds<--- another answer
t = .0721 Vo = 2.07 seconds
Why look at that, t = T/2. It spent just as much time going up as coming back down.
h = .0266 Vo^2 = 21.9 m <---- last answer