the formula: h(t)=1/2gt^2+vt+s where v is initial velocity and s is initial height. t = time g = 32 m/s.?

Question

A ball is thrown from a building with the above formula.
at 1.5 seconds, the ball reaches its apex 100ft above the ground
at 4 seconds it hits the ground.

Is there a way to find initial height and initial velocity without using the derivative?

Answer 1

The initial height is 64 feet, and the initial velocity is 48 ft/sec.

For the ball going up:

100 = 1/2 (-32)(1.5^2) + 1.5 v + s
1.5 v + s = 100 + 16 * 2.25 = 136 (Eq 1)

For the ball hitting the ground:

0 = 1/2 (-32)(4^2) + 4 v + s
4 v + s = 16 * 16 = 256 (Eq 2)

Subtract Eq 1 from Eq 2 to get:

2.5 v = 120 ==> v = 48 ft/sec (answer)

Then from Eq 1:

s = 136 - 1.5 * 48 = 64 feet (answer)

Answer 2

But calculus is your friend! Oh well...

I'm going to assume you meant g=-32 ft/s -- if its g=32m/s you are not on earth.

At t = 1.5, h = 100, so:

100 = -16(1.5)^2 + 1.5v + s
s = 136 - 1.5v

at t=4, h = 0, so:

0 = -512 +4v +s
s = 512 - 4v

so, solving those two equasions:

512 - 4v = 136 - 1.5v
2.5v = 376
v = 150.4

therefore:

s = 136 - 1.5(150.4)
s = -89.6

so, v = 150.4 ft/s, and s = -89.6 ft, assuming my above assumption about the acceleration is correct. If i'm wrong, you can follow my method to solve it anyway.

Answer 3

yes, you can make two equations and solve for two unknowns

you know at t=1.5 h(1.5)=100
and h(4)=0

this means that
100=1/2g*(1.5)62+v*1.5+s and
0=1/2g*(4)^2+v*4+s

Answer 4

Yes, of course!