Question

First timer help..

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

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

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

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)

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.