Average acceleration over a time interval lasting
is

where
is the difference in the jet's final and initial velocities. It's coming to a rest, so

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).
Answer:
A. 70 m OS the correct one
complete question:
An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot
Answer:
a ≈ 5281 ft
Explanation:
The observer at the top of a 462 ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°.
The angle of depression form the top of the cliff = 5°
The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°. 90° - 5° = 85° Note sum of an angle on a right angle is 90°.
using SOHCAHTOA principle we can solve for the distance from the base of the cliff to the point on the ground(a)
tan 85° = opposite / adjacent
tan 85° = a / 462
cross multiply
462 × tan 85° = a
a = 11.4300523 × 462
a = 5280.66 ft
a ≈ 5281 ft
Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t =
t =
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m