A projectile is fired into the air from the top of a cliff of height h = 208 m above a valley. Its initial velocity is v0 = 55.2 m/s at an angle θ = 54° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)

Do the whole vertical problem first

Vi = 55.2 sin 54
v = Vi - 9.81 t
at top v = 0
so
time to top = (55.2/9.81) sin 54
height at top = 208 + Vi t - 4.9 t^2
do that calculation
now you have t at top and height at top

now it falls from height H at top to ground. How long does that take?
0 = H - 4.9 t^2
t = sqrt (H/4.9)
that is time to fall so
total time in air = time to top + time to fall
NOW you can do the horizontal problem
u = 208 cos 54 the whole time
so
d = 208 cos 54 * total time in air

To find out where the projectile lands, we need to analyze its vertical and horizontal motion separately and then combine the results.

First, let's determine the time it takes for the projectile to reach the highest point of its trajectory. At the highest point, its vertical velocity will be zero. We can use the kinematic equation for vertical motion to find the time of flight to the highest point:

vf = vo + at,

where vf = 0 m/s (final vertical velocity), vo = 55.2 m/s (initial vertical velocity), and a = -9.8 m/s² (acceleration due to gravity). Rearranging the equation, we have:

0 = 55.2 - 9.8t,

9.8t = 55.2,

t = 5.63 s.

Since the projectile reaches the highest point halfway through the total time of flight, the total time of flight is twice the time to the highest point:

T_total = 2t = 2 * 5.63 s = 11.26 s.

Now, let's determine the horizontal displacement of the projectile. We can use the equation:

horizontal displacement = horizontal velocity * time,

where horizontal velocity is given by:

Vx = vo * cos(θ).

Substituting the given values:

Vx = 55.2 m/s * cos(54°) ≈ 55.2 m/s * 0.588 = 32.45 m/s.

Therefore, the horizontal displacement is:

horizontal displacement = 32.45 m/s * 11.26 s ≈ 365.64 m.

So, the projectile lands approximately 365.64 meters from the base of the cliff.