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

0.65 seconds

0.92 seconds

1.08 seconds

1.15 seconds

Verified Answer :

Greenleafable

Sharon
is jumping from an 18-foot diving board with an initial upward velocity
of 4 ft/s. When Susan jumps, Megan throws a beach ball up to Susan with
an initial upward velocity of 16 ft/s from a height 5 feet off the
ground. To the nearest hundredth of a second, it will take C. 1.08 seconds for the ball to reach Sharon after she jumps.

Answer : t = 1.08 seconds

Explanation :

Let \(x_1\) be the height of Sharon from ground, \(x_1=18\ foot\)

\(x_2\) be the height of Megan from ground, \(x_2=5\ foot\)

Initial upward velocity of Sharon, \(u_1=4\ ft/s\)

Initial upward velocity of Megan, \(u_2=16\ ft/s\)

Let h (t) is the height from ground when the ball reaches Sharon.

using the equation of motion as

\(h(t) =x_1+u_1t+\dfrac{1}{2}at^2\)

\(h(t)=18+4t+\dfrac{1}{2}\times 9.8t^2..................(1)\)

Similarly,

\(h(t)=5+16t+\dfrac{1}{2}\times 9.8t^2..................(1)\)

Equating equations (1) and (2)

\(4t+18=16t+5\)

\(t= 1.08\ s\)

The ball will reach after 1.08 seconds.

So, correct option is (c)

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