You fire
a projectile at 35 degrees above horizontal with an
initial velocity of 200 m/s
It lands in a Valley 300 m below
the launch point. What the time of flight?
Answers
Answer:
Time of flight: 25.78564632 sec or ≈ 25.8 sec
Range of projectile: 4224.472979 m or ≈ 4224.47 m
Solution:
Given: initial velocity ( vo ) = 200 m/s, initial height = 300 m, gravity(g) = 9.8 m/s 2 , and angle of launch ( θ ) = 35 ∘
Break down problem into two parts:
Launch of projectile above the horizon.(find: time and distance)
Projectile below the horizon. (find: time and distance)
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Launch of projectile above the horizon.(find: time and distance)
Find vox and voy (breaking down vo into x and y components).
vox=vo⋅cos(θ) ==> 163.8304089 m/s
voy=vo⋅sin(θ) ==> 114.7152873 m/s
Time of flight, for this part: Time(t) = vo⋅sin(θ)gravity(g)
t = 114.7152873/9.8 = 11.70564156 sec (this time is only the time to reach the top of projectile height). 11.70564156 sec x 2 = 23.41128312 sec
Distance(d) = r x t; r = vo⋅cos(θ)
d = vo⋅cos(θ) x t
d = 163.8304089 m/s x 23.41128312 sec = 3835.480085 m
2. Projectile below the horizon. (find: time and distance)
vox and voy will stay the same except voy which will be negative (this is because of -35 ∘ [below the horizon]).
vox=vo⋅cos(−θ) ==> 163.8304089 m/s
voy=vo⋅sin(−θ) ==> -114.7152873 m/s
vy=voy
Use of this little formula will do nicely: Distance(d) = vy⋅t−(1/2)⋅g⋅t2
d = height below horizon
300 = -114.7152873t - 4.9t 2
4.9t 2 + 114.7152873t - 300 = 0
Use this to solve for t: t = −b±b2−(4⋅a⋅c)√2⋅a
t = 2.374363203 sec and t = -25.78564632 sec (disregard)
Distance: s = t x ( vo⋅cos(θ) )
s = 2.374363203 sec x 163.8304089 m/s = 388.9928944 m
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
TOTAL TIME OF FLIGHT : ( part 1 and part 2) 2.374363203 sec + 23.41128312 sec = 25.78564632 sec or ≈ 25.8 sec
TOTAL DISTANCE TRAVELED: (part 1 and part 2) 3835.480085 m + 388.9928944 m = 4224.472979 m or ≈ 4224.47 m
Explanation:
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