46.2k views
1 vote
If an object is propelled straight upward from ground level with an initial velocity of 64feet per second, its height h in feet t seconds later is given by the equationh=- 16t2 +64t. After how many seconds is the height 48 feet?

User Gerard Yin
by
7.5k points

1 Answer

4 votes

After 1 sec and 3 sec, the height was 48 feet

Step-by-step explanation:

initial velocity = 64ft/sec

h= -16t² +64t

when h = 48ft

48 = - 16t2 +64t

-16t² +64t - 48 = 0

The above is a quadratic equation. So we will solve for t

Let's divide through by 16 as it is common to all the terms in the equation:

-t² + 4t - 3 = 0

taking the right side of the equation to the left side:

t² - 4t + 3 = 0

Using factorisation method:

factors: -3 and -1

t²-3t -t + 3 = 0

t(t - 3) -1(t - 3) = 0

(t-1)(t-3) = 0

t-1 = 0 or t-3 = 0

t =1 or t = 3

Inserting the values of t= 1 in the initial equation:

h= -16(1)² +64(1) = -16 + 64 = 48

Hence, t = 1 sec when the object is going up

This is because when an object moves up the direction is positive

Inserting the values of t= 3 in the initial equation:

h= -16(3)² +64(3) = -16(9) + 192 = -144 +192

h = 48

Hence, after 1 sec and 3 sec, the height was 48 feet

User KeithSmith
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories