Distance and time physics question?

Question

A stone is launched straight up by a slingshot. Its initial speed is 19 m/s and the stone is 1.5 m above the ground when launched. (a) How high above the ground does the stone rise? (b) How much time elapses before the stone hits the ground?

Answer 1

the equations for this are
v(t)=19-g*t for an object thrown vertically

and y(t)=1.5+19*t-.5*g*t^2

the stone will reach apogee when v(t)=0, so
0=19-g*t
solve for t
t=19/g
if g=10
t=1.9 seconds from launch to apogee
find y(1.9) by plugging into the equation
y(1.9)=1.5+19*1.9-5*1.9^2
or 19.55 m

then look at the time to descend to the ground
19.55=5*t^2
t=sqrt(19.55/5)
=1.98 seconds
add the two times together
1.9+1.98=3.88 s total flight time

j