gn.net/wely/Studient/Assigement-Reponses/submdep 13711011 (d) As it hit the hll,

Question

gn.net/wely/Studient/Assigement-Reponses/submdep 13711011 (d) As it hit the hll, what speed did it have and what angle dd its velocity make with the vertica? m/s 3 P309 Suppose a test pilot is able to safely of maneuvers, he is requred to fly the plane n a horizontal circle at its top speed of 1720 m/Th withstand an acceleration of up to S0 times the acceleration due to gravity (that s, remain conscious and alert enough to fly). During the course (a) What is the radius of the smallest aircle in which he will be able to safely fly the plane? lom (b) How long does it take him to go halifway around this mnimun-radus crce? Progress TOSHIBA

Explanation / Answer

1)

First convert miles per hour to metre per second

1720 mi/hr = 769 m/s

a)
a = v^2 / r
5g = (769)^2 / r
r = (769)^2 / 5g
r = 12069 m = 12.069 km

b)
Circumference = 2pi r
1/2 C = pi r
C = 12069 x 3.14 meters

C = 37897 m

distance = rate * time
time = distance / rate
t = 37897 m / 769 m/s
t = 49.28 s

2)

During the 20.2 seconds, the projectile rises 467 meters. Let’s use the following equation to determine the initial vertical velocity.
d = vi * t – 1/2 * a * t^2,
467 = vi * 20.2 - 1/2 * 9.8 * 20.2^2
vi = 122.1
This is approximately 122.1 m/s. This is vsin

Since the projectile moves a horizontal distance of 3170 meters in 20.2 seconds, let’s use the following equation to determine its initial horizontal velocity.

d = v*t
3170 = v*20.2
v = 3170/20.2
This is approximately 157 m/s. This is vcos . To determine the angle of the initial velocity, use the following equation.

tan = Vertical / Horizontal
tan = 122.1 / 157

= 37.87 degree


vf^2 = vi^2 + 2*a*d,

vf = 0, a = -9.8
0 = (122.1)^2 + 2*-9.8 * d
19.6 * d = (122.1)^2
d = 760.6 m

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