The distance traveled by a snowball thrown straight down from the top of a bluff can be calculated with the equation d=ut+12gt2, where d is the distance traveled in meters, u is the initial velocity in meters per second, t is the time in seconds, and g is the acceleration due to gravity, which is 9.8ms2. If the snowball traveled 200 meters in 5 seconds, what was its initial velocity?

15.5 m/s

Step-by-step explanation:

Using the equation, d = ut+1/2gt² where all the variables have meaning as given above, making u subject of the formula, we have

d = ut+1/2gt²

d - 1/2gt² = ut

d/t - 1/2gt²/t = u

u = d/t - 1/2gt

Since d = 200 m, t = 5 s and g = 9.8m/s², substituting the values of the variables into the equation, we have

u = d/t - 1/2gt

u = 200 m/5 s - 1/2 ×  9.8m/s² × 5 s

u = 40 m/s - 49 m/s ÷ 2

u = 40 m/s - 24.5 m/s

u = 15.5 m/s

So, the initial speed of the snowball is 15.5 m/s

c

step-by-step explanation:

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9