If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 112 ft/sec, its height after t seconds is s(t)=48+112t–16t^2. What is the maximum height the ball reaches "in ft"?What is the velocity of the ball when it hits the ground (height 0) "in ft/sec"?
You will need the vertex of this parabola
I don't know what method you learned , so I will use the simplest way not using Calculus
for y = ax^2 + bx + c, the
x of the vertex is -b/(2a)
so for yours,
t = -112/-32 = 3.5
sub that back in to get the maximum height.
when it hits the ground s(t) = 0
0 = 48 + 112t - 16t^2
16t^2 - 112t - 48 = 0
t^2 - 7t - 3 = 0
t = (7 ± √61)/2 = appr 7.4 seconds or a negative value of t, which we will reject.
To find the maximum height the ball reaches, we need to determine the vertex of the quadratic equation s(t) = 48 + 112t - 16t^2.
The vertex of a quadratic equation in the form s(t) = at^2 + bt + c is given by the formula t = -b / (2a). In this case, a = -16 and b = 112.
Substituting the values of a and b into the formula, we get t = -112 / (2 * -16) = 7/2 seconds.
To find the maximum height, we substitute this value of t into the equation:
s(7/2) = 48 + 112 * (7/2) - 16 * (7/2)^2
= 48 + 784/2 - 784/4
= 48 + 392 - 196
= 240 ft.
Therefore, the maximum height the ball reaches is 240 ft.
Next, to find the velocity of the ball when it hits the ground (height 0), we need to determine when s(t) = 0.
Setting s(t) = 0, we have:
48 + 112t - 16t^2 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. In this case, the equation factors as:
(t - 4)(-16t - 12) = 0
Setting each factor equal to zero, we find two possible solutions:
t - 4 = 0 => t = 4
-16t - 12 = 0 => t = -3/4
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the ball hits the ground after 4 seconds.
To find the velocity of the ball when it hits the ground, we can calculate the derivative of s(t) with respect to t, which represents the rate of change of height with respect to time:
v(t) = s'(t) = 112 - 32t
Substituting t = 4 into the equation:
v(4) = 112 - 32 * 4
= 112 - 128
= -16 ft/sec.
Therefore, the velocity of the ball when it hits the ground is -16 ft/sec. (Note: The negative sign indicates that the velocity is directed downwards.)