posted by Ali on .
A stone is launched straight up by a slingshot. Its initial speed is 20.1 m/s and the stone is 1.30 m above the ground when launched. Assume g = 9.80 m/s2.
(a) How high above the ground does the stone rise?
(b) How much time elapses before the stone hits the ground?
a. hmax = ho + (Vf^2-Vo^2 / 2g.
hmax=1.3 +(0-(20.1)^2 / -19.6=21.9 m.
b. Tr = Vf-Vo) / g,
Tr = (0-20.1) / -9.8 = 2.05 s. = Rise
time or time to reach max ht.
h = Vo*t + 0.5g*t^2 = 21.9 m.
0 + 4.9t^2 = 21.9,
t^2 = 4.47,
Tf = 2.11 s.Time to fall to ground.
Tr + Tf = 2.05 + 2.11 = 4.16 s. = Time