Question

Russel throws a soda from the balcony of his house to his friend Jones in the pool. If the soda is released 20 feet from the ground, and is thrown with a downward velocity of 2 ft/s, how long till the soda hits the pool, ignoring air resitance? Use the formula , where h is the height of the soda in feet and t is the time in seconds since Russel threw it. Round your answer to the nearest tenth.

A.

0.5 seconds B. 0.8 seconds C. 1.1 second D. 1.6 seconds

Answer 1

C. 1.1 seconds

Explanation:

The formula to use is :

`h=ut+(1)/(2)at^(2)`

Where: h is the height of 20 ft, u is the velocity of 2ft/s and a is the gravitational acceleration of 9.8m/s². The acceleration should be converted to ft/sec² to simplify calculations. The value is then 32.15ft/sec².

Substituting these values into the equation:

`20=2t+0.5*32.15t^(2) \\20=2t+16.075t^(2) \\0=16.075t^(2)+2t-20`

Using the quadratic formula, t can be solved.

a = 16.075, b = 2, c = -20:

`t=\frac{-b+-\sqrt{b^(2)-4ac} }{2a} \\t=\frac{-2+-\sqrt{2^(2)-4(16.075)(-20)} }{2(16.075)}\\t=(-2+-35.92)/(32.15)\\t=1.055 \\OR\\t=-1.179`

The answer cannot be negative, therefore the answer is t = 1.055 ≈ 1.1seconds

Answer 2

Using the formula S = ut +1/2at²
S=h= 20ft, u = 2ft/s, and a = 10m/s²
Thus; 20 =2t + 20t²
Therefore; 20t²+2t-20=0
Solving the equation quadratically for the value of t,
we get; t = 1.057 or t=-1.182
But t can not be negative,
Therefore, t= 1.1 seconds (to the nearest tenth)