Algebra question !?

Question

A ball is thrown into the air with an upward velocity if 36ft/s.
Its height is h in feet after t seconds is given by the function h= -16t^2+40t+6.

a. In how many seconds does the ball reach its maximum height?

b. What is the ball's maximum height?

c. When does it touch the ground?


Help me with this algebra question [:

Answer 1

a. To find the maximum height:

At the maximum height, the derivative of the height function h is zero (and the second derivative is less than 0).

the first derivative of h = dh/dt = -16*2t + 40.

Setting this to zero, we get the equation 32t = 40.

Therefore, t = 40/32 = 5/4 s. This is the time at which the first derivative of height is zero.

The second derivative of height is d2h/dt2 = -32. This is always negative. Implies, the t=5/4 is a maxima.

Therefore, the maximum height is reached at time t = 5/4seconds.

b. To get the max height, just substitute t = 5/4 in the h=-16t^2 + 40t + 6 equation.

hmax = -16*(5/4)^2 + 40*(5/4) + 6 = -25 + 50 + 6 = 31 ft.

c. The amount of time the ball takes to reach the top from the ground equals the time the ball takes to reach the ground from the top.

Therefore, the total time taken to touch the ground = time to top + time from top to ground = 5/4 + 5/4 = 5/2 s

Answer 2

Take the value of g as g = 10 ft/sec²
Question a.
h(t) = - 16t² + 40t + 6
h ` (t) = - 32 t + 40 = 0 at max. height
t = 40 / 32 = 5 / 4 secs.

Question b
h(5/4) = - 16 x 25/16 + 40 x 5/4 + 6
h(5/4) = - 25 + 50 + 6
h(5/4) = 31 ft (maximum height)

Question c
On downward journey:-
s = ut + (1/2).g.t² (but u = 0)
31 = 5 t²
t² = 31/5
t = 2.5 sec. (to 1 dec. place)

Total time = 1.25 + 2.5 sec
Total time = 3.75 sec

Answer 3

h(t) = -16t^2+40t+6

We look for the maximum of h(t) by taking the derivative h'(t) and setting it to 0.

h'(t)=-32t + 40=0

32t=40
t=40/32=5/4

Since h''(t)=-32<0 we know this is a maximum

if the maximum height is after 5/4 seconds then h(5/4) is the max height

h(5/4) = -16(5/4)^2+40(5/4)+6 = 31

Assuming ground level occurs when h(t)=0 then find the roots of h(t) and choose the result that makes sense (ie t>5/4)

Answer 4

a. Complete the square.
16 = 4^2
40 / 2 / 4 = 5
5^2 = 25

h = -16t^2 + 40t - 25 + 25 + 6
h = -(4t - 5)^2 + 31

Set 4t - 5 = 0
t = 5/4 = 1.25

The ball reaches it's maximum height after 1.25 seconds

b. h = -(4t - 5)^2 + 31
Setting the square part to 0, h = 31.

The max height is 31 ft.

c. When it touches the ground, the height is 0.
0 = -(4t - 5)^2 + 31
(4t - 5)^2 = 31
(4t - 5) = sqrt(31)
4t = sqrt(31) + 5
4t =~ 10.57
t =~ 2.64

The ball touches the ground after 2.64 seconds.

Answer 5

This is a parabola question. The maximum point is the vertex so you find it by using this (-b/2a,F(-b/2a)) So the maximum height would ge graphed at 5/4,31
5/4 would be the time it would take to be maximum. When does it touch ground? I don't know. Apparently I only learn mathematics. You can also derive the time by using calculus, but well since you're calculus one thats meaningless.
45 feet would have come to be without your equation. Well the height is 45 if the time was 5/4(by using your velocity. ) I'm guessing your equation was messed up? Or maybe I did.