Question
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) How high above the ground does the stone rise?
(b) How much time elapses before the stone hits the ground?
Answers
Henry
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
in flight.
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
in flight.