If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t)=32+80t–16t2. What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?
for max height I put 80-32t
and for velocity i put 32...which are wrong..
The building height is not the velocity.
For the max height, rewrite
s(t) = -16t^2 +80t +32 as
-16(t^2 - 5t + 25/4) +132
= -16(t -5/2)^2 + 132
The highest possible value of the function is 132 feet, and it occurs when t = 2.5 seconds.
For the velocity when it hits the ground, first solve for t when s = 0.
Then use the equation for velocity vs time:
V = 80 - 32 t ft/s
To find the maximum height reached by the ball, we need to determine the vertex of the quadratic function s(t) = 32 + 80t - 16t^2.
The vertex of a quadratic function in the form f(t) = at^2 + bt + c is given by the coordinates (t, f(t)), where t = -b / (2a). In this case, a = -16, b = 80, and c = 32.
Thus, the time at which the ball reaches the maximum height can be found by evaluating t = -80 / (2 * (-16)) = 80 / 32 = 2.5 seconds.
Now, to determine the maximum height, substitute the value of t into the equation s(t) = 32 + 80t - 16t^2:
s(2.5) = 32 + 80(2.5) - 16(2.5)^2
= 32 + 200 - 16(6.25)
= 32 + 200 - 100
= 232 feet.
Hence, the maximum height the ball reaches is 232 feet.
To find the velocity of the ball when it hits the ground (height 0), we need to determine the time when s(t) = 0.
0 = 32 + 80t - 16t^2
This equation is a quadratic equation, and we can solve it by factoring, using the quadratic formula, or by completing the square.
Let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 80, and c = 32.
t = (-(80) ± √((80)^2 - 4(-16)(32))) / (2(-16))
= (-80 ± √(6400 + 2048)) / -32
= (-80 ± √(8448)) / -32
= (-80 ± √(16 * 528)) / -32
= (-80 ± 4√528) / -32
= (-20 ± √528) / -8
Applying the quadratic formula, we get two solutions for t: t ≈ 0.228 and t ≈ 4.772.
Since the ball is thrown upward, the time when it hits the ground is t ≈ 4.772 seconds.
Finally, to find the velocity of the ball when it hits the ground, we differentiate the equation s(t) = 32 + 80t - 16t^2 with respect to t to get v(t), the velocity function.
v(t) = d(s(t)) / dt = d/dt(32 + 80t - 16t^2)
= 80 - 32t
Now, substitute t ≈ 4.772 into the velocity equation:
v(4.772) = 80 - 32(4.772)
= 80 - 152.704
≈ -72.704 ft/sec.
Hence, the velocity of the ball when it hits the ground is approximately -72.704 ft/sec.