posted by Zac .
Sally Sue is atop a 10.0 m high platform where she will drop an apple straight down. On the ground
below, and 20.0 m (horizontally) away, Betty Lou will shoot an arrow at the apple. If Betty's arrow has a
release speed of 25.0 m/s, at what angle must she fire that arrow, and at what time must Sally drop the
apple (relative to the firing of the arrow), such that the arrow will strike the apple at the highest point in
the arrow's trajectory?
H = 10 m is the initial height of the apple.
v(o)=25 m/s is initial velocity of the arrow
The angle at which the arrow starts is α .
h is the height at which the arrow strikes the apple (max height of the arrow).
We’ll use the expressions for the range, the height and the time of the projectile. You can find the derivation of these formulas in any textbook.
Horizontal x is the half of the projectile range
x=v(o)² •sin2α/2•g =>
sin2α =2•x•g/ v(o)² =2•20•9.8/25² =0.627,
2α =38.84º, α =19.42º.
The height of the projectile is
h= v(o)² •sin²α/2•g =625•0.11/2•9.8 =3.25 m.
The time that is needed for the arrow to rich the highest point is
t= v(o) •sinα/g=25•0.33/9.8 =0.84 s.
The apple coveres the vertical distance (H-h) for the time
t1 = sqrt(2•(H-h)/g) = sqrt(2(20-3.25)/9.8) = 1.84s.
Δt = t1-t=1.84 -0.84 = 1s.
Therefore, the arrow has to start its motion 1 s after the beginning of the apple motion.
I understand but you didn't put 10(initial height) into any of the equations?