another tough algebra question.?

Question

If an object is propelled straight upward from the ground level with an intial velocity of 64 feet per second its height h in feet t seconds later is giving by the equation h = -16t^2+64t

After how many seconds is the height 48 feet ?_________ Use comma to seprate the answers

Answer 1

u=initial velocity=64feet per sec
h=ut-1/2ft^2
-ve sign is due to upward movement
as the the given equation too implies
48=-16t^2+64t
or,16t^2-64t+48=0
diving this by 16 we get,
t^2-4t+3=0
or,t^2-3t-t+3=0
or,t(t-3)-1(t-3)=0
or,(t-3)(t-1)=0
so,t=3,1

Answer 2

Set 48 seconds equal to the function:
48=-16t^2+64t and factor
16(t^2-4t+3)
16(t-3)(t-1) soln: t = 3, t = 1
Now, put the solutions into the original equation:
1 = 48
2 = 64
3 = 48
4 = 0

This function describes the time of this projectile flight.
So, the answer is 1 second, the time necessary to reach 48 feet, in the upward trajectory as defined. It also reaches 48 feet in the downward trajectory.

Answer 3

48=-16t^2+64t You can simplify by 16
t^2 -4t+3=0 so t = ((4+-sqrt(4))/2

t= 1s (upwards) and t= 3s ( downwards)

Answer 4

48 = -16t^2 + 64t
0 = -16t^2 + 64t - 48

use quadratic fomula to solve for t