6 ft 200 ft FIGURE 2419 Arrow in Problem 54 55 An arrow is s

6 ft. 200 ft FIGURE 2.4.19 Arrow in Problem 54 55. An arrow is shot vertically upward with an initial velocity of 64 ft/s from a point above the ground. See FIGURE 2.4.20. (a) Find the height s(t) and the velocity v(t) of the arrow at time t e 0 (b) What is the maximum height attained by the arrow? What is the velocity of the arrow at the time the arrow attains its maximum height? (c) At what time does the arrow fall back to the 6-ft level? What is its velocity at this time? he height above ground of a toy rocket launched upward from the top of a building is given by s(t) 16t2 96t 56. (a) What is the height of the building? (b) What is the maximum height attained by the rocket? (c) Find the time when the rocketstrikes the round

Solution

Given that

S(t) =-16 t² +96t + 256

It is an equation of height of rocket at time t

(a)Initial postion of rocket is the height of the building because rocket is lanched from top of building.

Therefore initially time would be 0

Plug t=0 in equation in s(t) and we get

s(t) = 256

This is the height of building

Answer =256

(b) for maximum height we differentiate s(t) with respect to time t

d(s(t))/dt = -32t +96

Now for maximum height

-32t +96= 0

t =3 at this time rocket would attain maximum height

At t =3

Height s(t) = -16 *3² + 96*3 + 256

S(t) = 400

Answer

(C)

When rocket strike the ground at that time s(t) would be 0

Hence S(t) =0

-16t²+96t +256=0

Dividing by -16 on both side

t²-6t+16 =0

(t-8) (t+2) =0

t = 8 and t = -2

Here time t can\'t be negative

At t = 8 rocket would strike the ground

Answer: t =8