If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t)=48+80t–16t2. What is the maximum height the ball reaches? (my answer is 148). What is the velocity of the ball when it hits the ground (height 0)?
your first answer is correct.
for the second part, first find when the ball hits the ground, that is, set s(t) = 0
16t^2 - 80t - 48 = 0
t^2 - 5t - 3 = 0
t = (5 ± √37)/2 = 5.541 or a negative
sub 5.541 into your velocity, which you obviously had correct.
i did however my answer is still wrong. can you show me step by step. this way i can see what i am doing wrong. thank you
the answer's -97.32
To find the maximum height the ball reaches, we can look at the equation for height given by s(t) = 48 + 80t - 16t^2. The maximum height occurs at the vertex of the parabolic equation, so let's find the vertex.
The vertex of a parabola with equation y = ax^2 + bx + c can be found using the formula x = -b/(2a). In our case, a = -16, b = 80, and c = 48. Plugging these values into the formula, we get:
t = -80 / (2*(-16))
t = -80 / (-32)
t = 2.5
Now, let's find the corresponding height by plugging t = 2.5 into the equation:
s(2.5) = 48 + 80(2.5) - 16(2.5)^2
s(2.5) = 48 + 200 - 16(6.25)
s(2.5) = 48 + 200 - 100
s(2.5) = 148
So, the maximum height the ball reaches is indeed 148 feet.
Next, let's find the velocity of the ball when it hits the ground (at height 0). We can find the velocity by taking the derivative of the height function with respect to time (t):
v(t) = s'(t) = 80 - 32t
To find the velocity when the ball hits the ground, we set the height s(t) to 0 and solve for t:
0 = 48 + 80t - 16t^2
16t^2 - 80t - 48 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
t = (-(-80) ± sqrt((-80)^2 - 4(16)(-48))) / (2(16))
t = (80 ± sqrt(6400 + 3072)) / 32
t = (80 ± sqrt(9472)) / 32
Since we're interested in the time when the ball hits the ground, we take the positive root:
t = (80 + sqrt(9472)) / 32
Plugging this value of t into the velocity equation, we get:
v(t) = 80 - 32((80 + sqrt(9472)) / 32)
v(t) = 80 - 80 - 32sqrt(9472) / 32
v(t) = -32sqrt(9472) / 32
v(t) = -sqrt(9472)
Therefore, the velocity of the ball when it hits the ground is approximately -sqrt(9472) ft/sec.