Height of the Ball?

Question

The height s of a ball (in feet) thrown with an initial velocity of 60 feet per second from an initial height of 12 feet is given as a function of the time t (in seconds) by the equation
s(t) = -16t^2 + 60t + 12

(a) Find the time at which height is maximum?

(b) What is the maximum height?

Answer 1

s(t) = -16t^2 + 60t + 12

When ball is thrown up , height of the ball is maximum , when velocity is zero.

velocity, v is derivative of s.

ds/dt = -32t + 60

when this is zero

32t = 60

t = 60/32 = 15/8 sec

So the ball reaches maximum height when t = 15/8 s

substitute t = 15/8 in equation of s(t) to get the maximum height.

h(max) = s(15/8) = -16(15/8)^2 + 60(15/8) + 12

=> -225/4 + 450/4 + 12

=>273/4 = 68.25 ft

Answer 2

Part a)
s ` (t) = - 32 t + 60 = 0 for turning point
t = 60 / 32
t = 15/8
height is maximumum after 15/8 seconds

Part b)
s(15/8) = (-16)(225/64) + 900/8 + 12
s(15/8) = - 56.2 + 112.5 + 12
s(15/8) = 68.4
Max. height = 68.4 ft

Answer 3

that's out. A fulltoss isn't unlawful lower than the regulations of cricket and is not a rationalization for the umpire to call no ball, yet a bowler handing over too many finished tosses will be breaking the spirit of the game.

Answer 4

time to reach max when derivative is 0
ds/dt==32t+60=0
t=60/32=15/8=1 7/8second
height =-16(15/8)^2+60(15/8)+12
=68.25ft

Answer 5

v(t) = s'(t) = -32t + 60
-32t + 60 = 0
t = 60/32

Maximum height:
s(60/32) = -16(60/32)^2 + 60(60/32) + 12

Answer 6

ds/dt = -32t + 60 = 0
t=60/32

Answer 7

set derivative equal to 0, solve for t.
Plug this t into s(t) for the max height.