AP Physics Question?

Question

The determined coyote is out once more in pursuit of the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 13.0 m/s2 (Fig. P4.65). The coyote starts at rest 70.0 m from the brink of a cliff at the instant the roadrunner zips past him in the direction of the cliff.

Figure P4.65

(a) The roadrunner moves with constant speed. Determine the minimum speed he must have so as to reach the cliff before the coyote.
m/s
(b) At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote's skates remain horizontal and continue to operate while he is in flight so that his acceleration while in the air is (13.0 - 9.80) m/s2. The cliff is 100 m above the flat floor of a wide canyon. Determine where the coyote lands in the canyon.

Your answer differs from the correct answer by 10% to 100%. m (from base of cliff)
(c) Determine the components of the coyote's impact velocity. (Assume right is the positive x direction and up is the positive y direction.) vx = m/s
vy = m/s

Answer 1

The roadrunner has
x(t)=70-v*t

and the coyote has
v(t)=13*t
and
x(t)=70-7.5*t^2

a) When x(t)=0, both arrive at the same instant
70-7.5*t^2=0
t=sqrt(70/7.5)
70-v*t=0
v=70/sqrt(70/7.5)
v>=23

b) Once airborne, I will assume the coyote keeps his skates horizontal, and I will assume for this segment the take-off time is t=0;
x(t)=39.7*t+7.5*t^2
y(t)=100-.5*9.81*t^2
when the coyote impacts, his y(t)=0, solve for t
t=sqrt(200/9.81)
t=4.515 seconds
x(4.515)= 332 m

c)
vx=39.7+13*4.515
vx=98.4 m/s

vy=-9.81*4.515
vy=-44.3

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