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A ball is thrown straight upward. The position of the ball is s(t)=-16t^(2)+8t+3 where s is the distance from the ground to the ball in feet an How high is the ball after (1)/(4) second? At what time

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The ball's position as a function of time is given by the equation s(t) = -16t^2 + 8t + 3, where s denotes the height in feet from the ground and t is time in seconds.

We can substitute t = 1/4 in the given equation to find the height of the ball after 1/4 second.

So, s(1/4) = -16*(1/4)^2 + 8*(1/4) + 3 = 4 feet

Now, let's calculate the time when the ball reaches its highest point.

The highest point is reached when the velocity of the ball becomes zero (as the ball stops for a moment before it starts descending). To find when this happens, we need to first find the velocity function, v(t), which is the derivative of the position function, s(t).

The derivative of s(t) is v(t) = -32t + 8.

We can set v(t) equal to zero and solve for t to find the time at which the ball reaches its highest point.

So, -32t + 8 = 0. Solving this gives us t = 1/4 second.

So, the height of the ball 1/4 second after it is thrown is 4 feet, and the ball reaches its highest point 1/4 second after it is thrown.

User Shahjahan Ravjee
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