Question

First timer help..

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

Answer 1

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+(1)/(2)at^2`

`h(t)=18+4t+(1)/(2)* 9.8t^2..................(1)`

Similarly,

`h(t)=5+16t+(1)/(2)* 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)

Answer 2

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.