If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 16 ft/sec, its height after t seconds is s(t)=64+16t−16t^2.

what is the maximum height the ball reaches?

What is the velocity of the ball when it hits the ground (height 0)

Question

If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 16 ft/sec, its height after t seconds is s(t)=64+16t−16t^2. 

what is the maximum height the ball reaches?

 What is the velocity of the ball when it hits the ground (height 0)

anonymous · Answer

hey notsobright can you help me with another one after please

anonymous · Answer

Take derivative and equate it to zero
16-32t=0
t=1/2 
h=64+ 8-4
h=68

anonymous · Answer

Yeah sure

anonymous · Answer

ok

anonymous · Answer

what is the slope of the normal line of m2
the equation of the normal lineis y=?

y=10cosx at the point (π/3,5).
derivative of y is -10sinx

The slope of the tangent line is m1 is -10sin(pi/3)
the equation of the tangent line is y= -5sqrt(3)x+(5pi/sqrt3)+5

anonymous · Answer

Rephrase it it is not clear

anonymous · Answer

ok

anonymous · Answer

Find equations of the tangent line and normal line to the curve y=10cosx at the point (π/3,5). 

Derivative of y= -10sinx
The slope of the tangent line is m1=-10sin(pi/3)
The equation of the tangent line is y=-5sqrt(3)x+(5pi/sqrt3)+5

The slope of the normal line is m2=?
The equation of the normal line is y=?