How do I find the velocity of the rock when it was initially thrown?

Question

A rock is thrown into the air from atop a building, and its height at time t is given by the equation

h(t)= -16t^2+48t+60

What was the velocity of the rock when it was initially thrown?

Any help would be greatly appreciated. Please show your work,Thanks!

Answer 1

48 feet per second = the initial velocity

-26T62 is the gravity term

48t is the initial velocity term where the coefficient of t is the initial velocity

60 is the initial height

it's written in standard form h(t) = (a/2)t^2 + vot + ho

where a is the acceleration or deceleration term if negative, IF it's gravity then a =-32 meaning it decelerates 32 feet per second per second

vo is initial velocity

ho is initial height

h'(t) = -32t + 48=0

t = 48/32 = 6/4 = 1.5 seconds when the rock reaches maximum height

max h(1.5) = - 16(1.5^2) = 48(1.5) + 60

= -16(2.25) + 72+60

= -36 + 132 = 96 feet

the other question that comes up in this type of problem is when does it hit the ground

just set h =0 and solve for t

-16t^2 + 48t +60 =0

16t^2 -48t = 60

t^2 -3t = 60/16 = 15/4

t^2-3t + 9/4 = 15/4+ 9/4 = 24/4 = 6

(t-3/2)^2 = 6

t-3/2 = sqr6 = about 2.45

t = 1.5 + 2.45 = 3.95 seconds