Question

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?

Answer 1

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