hw help needed !

Question

I drop a ball off a three-story balcony. We know that objects in freefall follow the equation for height h, in meters, h=−4.9t2+ch=-4.9t2+c, where t is time in seconds, and c is the initial height. A student timed the drop at 1.26 seconds.

Use this to determine the height from which the ball was dropped, to at least 2 decimal places.

Answer 1

 h(t)=−4.9t²+c

We know know t=1.26 thats how long it took the ball to hit the ground and the ground is 0 feet

so we have 0 = -4.9(1.26)² + c now solve for c

c = 7.78

so we have h(t)=−4.9t²+ 7.78 now to determine the height t = 0

h(0)=−4.9(0)²+ 7.78

h(0) = 7.77 this says in 0 seconds the ball is at 7.78 feet from the ground

Answer 2

Let's do the problem knowing that it is a standard physics problem on free fall. It is given to us that

h = c - 4.9t²

Here h - height at any given moment,

c = h₀ - initial height. i.e. the height at time 0.

When the body reached the ground, then h = 0. It means , we can write:

0 = c - 4.9t²

From this equation

c = 4.9t²

Putting t = 1.26 s into the equation, we will have for c:

c = 4.9(1.26)² = 7.78 m

And this is the final answer: 7.78 m

Answer 3

h(t) = -4.9t^2 + c

t= 1.26

h(1.26) = -4.9(1.26)^2 + c = 0

c= 4.9(1.26)(1.26)

c= 7.77924 meters = intial height

general equation is:

h(t) =-4.9t^2 + vot + ho

where vo=initial velocity = 0 if it is just dropped

ho = initial height = c = the constant term measured in meters

t= time

h(t) = height at time t in seonds

-4.9t^2 is the acceleration due to gravity. It's accelerating at -9.8 meters per second per second. that's the change in velocity. at time t=0 velocity=0, in 1.26 seconds the velocity= -9.8(1.26) meters per second

h'(t) = v(t) = the derivative of h(t)= -9.8t