A, B, and C are relatively straightforward. D is somewhat more complex.
A. The height, y, is given by
y = yo + vo t + (1/2) a t^2
Where (SI units)
yo = initial height = 0 m
vo = initial velocity = 60 m/s
a = acceleration of gravity = -10 m/s^2 approx
t = elapsed time, s
We get
y = 0 + 60 t - 5 t^2
At the top of the trajectory, the kinetic energy is 0, and the potential energy is m g h,
and it matches the initial kinetic energy, (1/2) m (vo)^2
m g h = (1/2) m (vo)^2
or
h = (vo)^2 / 2 g
h = (3600) / (20) = 180 m
good idea to check units
[m] = [m^2/s^2] / [m/s^2] = [m] ? yes
B. The time in the air to the apex can be gotten by solving for when the velocity becomes 0:
v = vo - g t
v = 60 - 10 t
0 = 60 - 10 t
t = 6 s to the apex
Check:
use t = 6 and find height
h = 0 + (60)(6) - (5)(6)^2 = 0 + 360 - 180 = 180 m yes
Arguably the time coming down is the same duration as the time going up,
for a total time of t =12 s.
Alternatively, can also start with
y = yo + vo t + (1/2) a t^2
And go from y = 0 to y = 0 return
0 = 0 + 60 t - 5 t^2
5t^2 = 60 t and divide both sides by t and simplify, giving
t = 12 s
Could also solve this using the standard quadratic equation form
a x^2 + b x + c = 0 with the solution
x = [- b +- sqrt(b^2-4a c) / 2 a, etc.
This will give you two solutions (note the +-) for t, one for on the way up and the other for on the way down.
C. The initial velocity for the second arrow is gotten from its maximum height
m g h = (1/2) m (vo)^2
2 g h = (vo)^2
(2)(10)(46) = (vo)^2
vo = 30. m/s
Check height at apex
0 = vo - g t
t = vo / g = 3.0 s
h = 0 + (30.) (3.0) - (0.5) (10) (3.0)^2 = 45 m (close enough)
D. Is more complex, and I am leaving it as "an exercise for the student."
Let y1 be the trajectory of the first arrow as a function of t1. Solve t1 for h=46 on the way down.
Let y2 be the trajectory of the first arrow as a function of t2. Solve t2 for h=46 at the apex.
Determine the time difference between t1 and t2 as the delay time for launching arrow 2.
