Jo throws a ball from a point 40 m above the ground. The height of the ball t seconds after it's thrown can be modeled by the equation f(t) = 40t - 5t^2 What is the max height the ball will reach?

Question

And I also need to know how long it will take the ball to hit the ground I kinda need this answer within an hour if someone could get back to me quickly that would be great

Answer 1

f(t) = 40t -5t^2

take the derivative and set equal to zero, then solve for t=time in seconds

f'(t) = 40-10t = 0

10t = 40

t = 4 seconds to reach maximum height

plug t=4 into the original equation to solve for maximum height

max height = f(4) = 40(4) -5(4)^2

f(4) = 160 -80

= 80 meters

it hits the ground in twice as much time, 2t = 2(4) = 8 seconds

or set f(t) = 0 and solve for t. zero height is ground level

40t - 5t^2 = 0

factor

5t(8 - t) = 0

set each factor equal to zero and solve for t

5t = 0

t = 0/5 = 0

8-t = 0

t = 8

t = 0 or 8 seconds when it's at ground level.

It starts at ground level and returns 8 seconds later