Velocity/ max height question?

Question

ok I came up with the answer 36ft but I do not think this is correct, please help me with the steps I should take to resolve this.

"if a ball is thrown straight up with an initial velocity of 48 ft per sec., its height after t seconds is given by the formula s=48t-16tsquared. Find the maximum height attained by the ball."

Answer 1

Velocity = g * t, so t = v / g = 48 / 32 = 1.5 seconds.

Distance = 1/2 g t^2 = 16 * 2.25 = 36 feet.

Answer 2

Its 36 feet

S= 48t - 16t^2. Using Calculus, you can find the maximum hight reached by differentiating S and finding the value for t when the derivative = 0 >>> 48 - (16x2)t = 0 >>> t = 48/32 = 1.5

The maximum S (height) is when t (time) = 1.5 seconds. Then Plug 1.5 seconds to find where the ball lies at 1.5 seconds. S(1.5)= 48(1.5) - 16 (1.5)^2 = 36 feet

Answer 3

The maximum height occurs when t = -48/(2(-16))

or t = -48/-32 = 1.5 seconds

Plug this in for t to get the maximum height.

It does come out 36 ft so you are right.

PS If you are supposed to use calculus instead of algebra, see Dupinder's answer below. She is correct.

Answer 4

I also got 36 feet.

The graph of the position of the ball is a parabola opening down. The maximum point of such a parabola is at its vertex. The equation for the x value of the vertex is -b/2a. a=-16, and b=48 in your equation, so the x value equals -(48)/2(-16), or 1.5. Plug in 1.5 into the equation, and the maximum height equals 48(1.5)-16(1.5)^2 -> 72-36 -> 36.

Answer 5

v^2=u^2+2as the position, v=very last speed u=initial speed a=acceleration (through gravity hence) s=distance if you're calculating from floor element, v=0 (at max height). if you're calculating from max height, u=0. Substituting provide you with a max speed of two hundred m/s.

Answer 6

s=48t-16t^2
diff wrt x
ds/dt=48-32t
at max. height
ds/dt=0
so
0=48-32t
t=1.5
put this in
s=48t-16t^2
and find s