Solved Problems in Linear Motion – Down motion in free fall
1. A ball is thrown vertically downward with initial speed 10 m/s and reach the ground in 2 seconds. Find final speed just before the ball hits the ground. Acceleratio gravitatis (g) = 10 m/s2Resistentiam aeris neglege.
Notum:
Velocitas initialis (vo) = 10m/s
Time elapsed (t) = 2 seconds
Acceleratio gravitatis (g) = 10 m/s2
Wanted : Final velocity (vt)
solution:
Acceleratio 10 m/s2 means speed increase by 10 m/s each second. After 3 second, speed = 30 m/s.
Final velocity = 10 m/s + 20 m/s = 30 m/s.
Kinematic equations for motus cum acceleratione constanti, as shown below :
vt v =o + a t ………. 1
h = vo t + ½ apud2 ………. 2
vt2 v =o2 + 2 a h ………. 3
vt v =o + gt
vt = 10 + (10)(2)
vt = 10 + 20 = 30 m/s
Final velocity = vt = 30m/s
2. A stone is thrown vertically downward from a bridge with initial speed 5 m/s and reach the water in 2 seconds. Calculate the height of the bridge.
Notum:
Velocitas initialis (vo) = 5m/s
Time elapsed (t) = 2 seconds
Acceleratio gravitatis (g) = 10 m/s2
Quaesitum: the height of the bridge (h)
solution:
h = vo t + ½ gt2
h = (5)(2) + ½ (10)(2)2
h = 10 + (5)(4)
h = 10 + 20
h = 30 metra
3. A ball is thrown vertically downward with initial speed 10 m/s from a height of 80 meters. Find (a) Time in air (b) Final velocity just before ball strikes the ground.
Notum:
height (h) = 80 meters
Velocitas initialis (vo) = 10m/s
Acceleratio gravitatis (g) = 10 m/s2
Quaesitum:
(a) Intervallum temporis (t)
(b) Velocitas finalis (v)t)
solution:
(a) Intervallum temporis (t)
Final velocity :
vt2 v =o2 + 2 gh
vt2 = (10)2 + 2(10)(80) = 100 + 1600 = 1700
vt = 41m/s
Time interval (t) :
vt v =o + gt
41 = 10 + (10)(t)
41 – 10 = 10 t
31 = 10 t
t = 31 / 10 = 3,1 secunda
(b) Velocitas finalis (v)t) ?
vt = 41m/s
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