contestada

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

Respuesta :

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 [tex]x_1[/tex] be the height of Sharon from ground, [tex]x_1=18\ foot[/tex]

[tex]x_2[/tex] be the height of Megan from ground, [tex]x_2=5\ foot[/tex]

Initial upward velocity of Sharon, [tex]u_1=4\ ft/s[/tex]

Initial upward velocity of Megan, [tex]u_2=16\ ft/s[/tex]

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

using the equation of motion as

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

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

Similarly,

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

Equating equations (1) and (2)

[tex]4t+18=16t+5[/tex]

[tex]t= 1.08\ s[/tex]

The ball will reach after 1.08 seconds.

So, correct option is (c)

Otras preguntas