Question

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t² + 20t + 65. What is the object's maximum height? the numeric answer only, rounded to the nearest meter.

Answer 1

Maximum height will be 85 meter

Step-by-step explanation:

given that height of an object is function of time and it is given as

`h(t) = -4.9 t^2 + 20t + 65`

here we can write it as

`h(t) - 65 = 20 t + (1)/(2)(-9.8) t^2`

now we can compare it with the kinematics equation

`y(t) - y_o = v_i t + (1)/(2)at^2`

now if we compare the two equations then we will have

initial height of object = 65 m

initial velocity of projection = 20 m/s

acceleration = - 9.8 m/s/s

now for the maximum height at which its final speed will become zero is given as

`v_f^2 - v_i^2 = 2a(h - h_o)`

`0 - 20^2 = 2(-9.8)(h - 65)`

`-400 = -19.6(h - 65)`

`h = 85.4 m`

So maximum height in meter in nearest integer will be 85 m

Answer 2

To answer the question above,first you have to take the derivative of the h(t) function and set it equal to zero , then solve for t to find the time when the object is at its maximum height :

h ' (t) = - 9.8 t + 20

- 9.8 t + 20 = 0

- 9.8 t = - 20

t = ( - 20 / - 9.8 ) = 2.04 seconds

(I will round that to 2.0 seconds, which is acceptable because of the significant figures used in the question, but I won't go into that.)

So the object reaches max. height at t = 2

Put that value of t into the original h(t) function to find the height at time t = 2 :

h(2) = ( - 4.9 * 2 ) + ( 20 * 2 ) + 65

h(2) = ( - 9.8 + 40 + 65 ) = 95.2 meters.