Question

Sharon is jumping from an 18-foot diving board with an initial upward velocity of 4 ft/sWhen Sharon jumps, Megan throws a beach ball up to Sharon with an initial upward velocity of 16 ft/s from a height 5 feet off the groundTo the nearest hundredth of a second, how long after she jumps does the ball reach Sharon?

Answer 1

It's C: 1.08 seconds

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Answer 2

1.08 seconds

Explanation:

given data

first equation:

According to the data, sharon position will be

h(t)= -16
`t^(2)` + 4t +18---->eq(1)

Second equation;

When Sharon jumps, Megan throws a beach ball up to Sharon with an initial upward velocity of 16 ft/s from a height 5 feet off the ground to the nearest hundredth of a second

therefore, ball position would be

h(t)= -16
`t^(2)` +16t +5 --->eq(2)

In order to find how long after she jumps does the ball reach Sharon i.e time, we will equate eq(1) and eq(2)

Therefore,

eq(1)= eq(2)

-16
`t^(2)` +4t +18 = -16
`t^(2)` + 16t +5 --->(further simplifying it)

-16
`t^(2)` + 16
`t^(2)` + 18 -5 = 16t - 4t

13 = 12t

t= 13/12

t= 1.08 s

Thus it took 1.08 seconds.