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borishaifa [10]
3 years ago
9

The projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is

the time in seconds since launch, s(t) is the height after t seconds, v0 is the initial velocity, and s0 is the initial height. The acceleration due to gravity is –4.9 m/s2. An object is launched at an initial velocity of 19.6 meters per second from an initial height of 24.5 meters. Which equation can be used to find the number of seconds it takes the object to hit the ground? 0 = –4.9t2 + 19.6t + 24.5 0 = –4.9t2 + 24.5t + 19.6 19.6 = –4.9t2 + 24.5t 24.5 = –4.9t2 + 19.6t
Mathematics
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0

Answer:

I think its A on edge

Step-by-step explanation:

GREYUIT [131]3 years ago
5 0
So for projectile motion questions, we use the Cartesian equations.
Given that s(t) is the height of the projectile at an arbitrary time, t, we want the height of the projectile to equal to zero in order to find its flight time.

Let s(t) = 0.
gt^{2} + V_0 t + s_0 = 0
Now, the initial velocity of the projectile is given to be 19.6, and the initial height is 24.5.

Substitute those in to find:
-4.9t^{2} + 19.6t + 24.5 = 0
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In triangle ABC, angle A=50°, angle C=65°. Point F is on AC such that, BF is perpendicular to AC. D is a point on BF (extended)
LiRa [457]

Answer:

The length of EF is 5.29

Step-by-step explanation:

If BC = 12 then;

FB = 12 × sin65 = 10.876

BA = FB/(sin50) = 10.876/(sin50) = 14.2

BD = AB (Sides of isosceles triangle) = 14.2

∴ DF = BD + FB = 10.876 + 14.2 = 25.07

∠F C D = tan⁻¹(D F/F C) =  tan⁻¹(25.07/(12×cos65)) = 78.6

∠F A D = tan⁻¹(D F/A F) =  tan⁻¹(25.07/(14.2×cos50)) = 70

∴ ∠D = 180 - (78.6° + 70°) = 31.4°

AE = AD × sin31.4° = 2×AB×cos(70° - 50°) × sin31.4

∴ AE =  2×14.2×cos(20°) × sin31.4 = 13.92

EC = AC × cos78.6° = (AB×cos50 + BC×cos65)×cos78.6° = 2.815

Therefore, from cosine rule, a² = b² + c² - 2·b·c × cos(A)

Hence, we have;

EF² = EC² + FC² - 2 × EC × FC ×cos∠FCD (Cosine rule)

That is, EF² = 2.815² + (12×cos65)² - 2 × 2.815× (12×cos65) ×cos78.6°

EF² = 27.98

∴EF = √27.98 = 5.29.

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3 years ago
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schepotkina [342]
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An amount of $41,000 is borrowed for 7 years at 7.5% interest, compounded annually. If the loan is paid in full at the end of th
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I=prt
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