Roscoe hit a ball straight up at a speed of 96 ft/s. His bat hit the ball at a height 3 ft above the ground. After how many seconds did the ball hit the ground? (Give your answer to the nearest hundredth of a second.)

Answer:

t = 6.031 seconds

Step-by-step explanation:

Roscoe hit a ball straight up at a speed of 96 ft/s.

His bat hit the ball at a height 3 ft above the ground.

We need to find how many seconds did the ball hit the ground.

Using equation of motion to find height of the ball.

[tex]h=-16t^2+96t+3[/tex]

When it hits the ground.

h = 0

[tex]-16t^2+96t+3=0[/tex]

It is a quadrctic equation where a = -16, b = 96 and c = 3

[tex]t=\dfrac{-96\pm \sqrt{96^2-4(-16)(3)} }{2(-16)}\\\\t=-0.031\ s, 6.031\ s[/tex]

Neglecting negative value.

So, after 6.031 seconds the ball will hit the ground.

Given :

Roscoe hit a ball straight up at a speed of 96 ft/s.

His bat hit the ball at a height 3 ft above the ground.

To Find :

After how many seconds did the ball hit the ground.

Solution :

When the ball reaches the same height from which it is thrown ( i.e. 3 ft ) its displacement will be zero.

So, using equation of motion :

[tex]s = ut + \dfrac{at^2}{2}\\\\96t - \dfrac{10t^2}{2} = 0\\\\5t^2 - 96t = 0\\\\t( 5t - 96 ) = 0\\\\t = \dfrac{96}{5}\\\\t = 19.2 \ s[/tex]

Now, time taken to reach ground from 3 feet of height is :

[tex]s = ut + \dfrac{at^2}{2}\\\\3 = 96t + 5t^2 \\\\5t^2 + 96t - 3 = 0[/tex]

t = 0.03 s

Therefore, time taken is T = 19.2 + 0.03  = 19.23 seconds.

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