Answer:
665 ft
Explanation:
Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.
The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is
Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees
Since the 2 sides length above make up the 200 foot monument, their total length is
0.231d + 0.07d = 200
0.301 d = 200
d = 200 / 0.301 = 665 ft
It is typically 30 km to 50 km thick.
Answer:
The people with caculators will probably answer faster due to thier ablitiy to use a device of technology
Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where 
is the initial velocity.
(a).
When the projectile hits the 50m mark, 
; therefore,
solving for 
we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed 
of
