An object is thrown into the air has height given by h(t)= -16t^2 + 64t + 60feet, t seconds after it is thrown.

Question

An object is thrown into the air has height given by h(t)= -16t^2 + 64t + 60feet, t seconds after it is thrown.

The object reaches the highest point _____ seconds after it is thrown

At that time the objects height is ____ feet

Answer 1

Hi Joan

h(t)= -16t^2 + 64t + 60

This can be viewed and treated as a parabola in the standard form

f(x) = ax² + bx + c

where x = t

h(t) = f(x) or y

a = -16, b = 64 c= 60

When a is negative the parabola opens downward so the coordinates of the vertex, give the time when the maximum height is reached.

The x coordinate of the vertex is -b/2a

-64/2(-16) = -64/-32 = 2 so t is 2 seconds

We just plug in x = t = 2 to find the y or height coordinate of the vertex

f(2) = -16(2)² + 64(2) + 60

f(2) = -16(4) + 128 + 60

f(2) = -64 + 128 + 60

f(2) = 64 + 60

f(2) = 124, the y or height coordinate of the vertex.

Answer 2

To find the highest point where the object is thrown, we need to look for the vertex of the quadratic function,

h(t) = -16t² + 64t + 60, where t = time (in seconds) and h = height (in feet).

t = -b/2a

t = -64/2(-16) = -64/(-32) = 2 seconds

Plug in 2 for t to find the object's height after being thrown.

h(2) = -16(2)² + 64(2) + 60 = -16(4) + 64(2) + 60 = -64 + 128 + 60 = 124 feet

The object reaches the highest point at 2 seconds after it is thrown. At that time the object's height is 124 feet.

Answer 3

in 2 seconds it reaches max height of 124 feet

take the derivative & set equal to zero, solve for t, then substitute back into the original equation to get h(2)

h' = -32t + 64= 0

t = 64/32 =2

-16(2)^2 + 64(2) + 60 = 124 feet

Answer 4

Joan

Draw the function y = -16x² + 64x + 60

on Desmos or a graphing calculator, then you can just read your answers off the graph.