If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 112 ft/sec, its height after t seconds is s(t)=32+112t–16t2 . What is the maximum height the ball reaches?What is the velocity of the ball when it hits the ground (height 0 )?
v(t) = s'(t) = 112 - 32t
at max v = 0
32t = 112
t = 3.5
then s(3.5) = ......
when it hits the ground, s(t) = 0
16t^2 - 112t - 32 = 0
t^2 - 7t - 2 = 0
solve using the quadratic formula,
sub the positive answer into v(t)
Hi I have the same question. The first part is correct to find the maximum height, but while solving for the quadratic formula to find the velocity when it hits the ground, the answer is incorrect. (You would get 7.2749 and -0.27492 after solving the quadratic)
Nevermind! I figured out you're supposed the plug the positive value into the derivative v'(t).
To find the maximum height the ball reaches, we need to determine the highest point on the graph of the equation s(t) = 32 + 112t - 16t^2. This represents the height of the ball as a function of time.
The equation s(t) represents a downward-opening quadratic curve because of the negative coefficient for the t^2 term (-16t^2). The highest point on this curve is known as the vertex.
The vertex of a quadratic function can be determined using the formula t = -b/2a, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c.
For the equation s(t) = 32 + 112t - 16t^2, the coefficient of the t^2 term is -16, so a = -16, and the coefficient of the t term is 112, so b = 112.
Now, we can find the t-coordinate of the vertex using t = -b/2a.
t = -112 / (2 * -16) = -112 / -32 = 3.5.
Therefore, the maximum height is reached at t = 3.5 seconds.
To find the maximum height reached by the ball, substitute this value of t back into the equation s(t):
s(3.5) = 32 + 112(3.5) - 16(3.5)^2.
Calculating this expression gives us:
s(3.5) = 32 + 392 - 196 = 228 feet.
Therefore, the maximum height the ball reaches is 228 feet.
To find the velocity of the ball when it hits the ground (height 0), we need to find the time it takes for the ball to reach the ground. Since the height of the ball when it hits the ground is 0, we can set s(t) = 0 and solve for t.
0 = 32 + 112t - 16t^2.
By rearranging this equation, we get:
16t^2 - 112t - 32 = 0.
Simplifying further by dividing every term by 16:
t^2 - 7t - 2 = 0.
This is a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 1, b = -7, and c = -2.
Substituting these values into the quadratic formula:
t = (-(-7) ± √((-7)^2 - 4(1)(-2))) / (2(1)).
Simplifying further:
t = (7 ± √(49 + 8)) / 2.
t = (7 ± √57) / 2.
The two possible solutions for t are:
t = (7 + √57) / 2 ≈ 6.72 seconds,
or
t = (7 - √57) / 2 ≈ 0.28 seconds.
Since the ball was thrown upward, we need the positive solution. Therefore, the time it takes for the ball to hit the ground is approximately 6.72 seconds.
Now, to find the velocity of the ball when it hits the ground, we can differentiate the equation s(t) with respect to t to obtain the velocity function v(t).
v(t) = s'(t) = 112 - 32t.
Substituting t = 6.72 into the velocity function:
v(6.72) ≈ 112 - 32(6.72) ≈ 112 - 214.4 ≈ -102.4 ft/sec.
The negative sign indicates that the ball is moving downward, which is consistent with its vertical direction.
Therefore, the velocity of the ball when it hits the ground is approximately -102.4 ft/sec.