1. A roadrunner runs directly off a cliff with an initial velocity of 3.5m/s.
A) What are the components of this velocity?
B) What will be the horizontal velocity 2 seconds after the bird leaves the cliff?
C) If the cliff is 300m high, at what time will the roadrunner reach the ground?
D) How far from the cliff will this bird land?
E) If there is a small pond which begins 25m away from the cliff and extends 2.5m from there; will the roadrunner land in the pond? Explain.
F) What is the final velocity(Vy) at which the roadrunner is traveling? [The vertical velocity at be time when the bird reaches the ground]
G) What is the final horizontal velocity(Vx) at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]
H) What is the total final velocity of this motion? [Magnitude (V^2 = Vy ^2 + Vx^2 and direction theta =tan^-1 (Vy/Vx)]
A) What are the components of this velocity?
B) What will be the horizontal velocity 2 seconds after the bird leaves the cliff?
C) If the cliff is 300m high, at what time will the roadrunner reach the ground?
D) How far from the cliff will this bird land?
E) If there is a small pond which begins 25m away from the cliff and extends 2.5m from there; will the roadrunner land in the pond? Explain.
F) What is the final velocity(Vy) at which the roadrunner is traveling? [The vertical velocity at be time when the bird reaches the ground]
G) What is the final horizontal velocity(Vx) at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]
H) What is the total final velocity of this motion? [Magnitude (V^2 = Vy ^2 + Vx^2 and direction theta =tan^-1 (Vy/Vx)]
-
A.) The components are 3.5m/s horizontally, and 0m/s vertically.
B.) Horizontal velocity is constant, so it will still be 3.5m/s, this is the same for G.
C.) For vertical, use d=vi(t)+.5(a)(t^2), vi=o, a=9.8, d=300m, t=?, so t=7.8s.
D.) Constant velocity horizontally, so use d=vt, so d=3.5(7.8), so d=27.3m.
E.) Yes, becasue it falls in to this, becasue it extends from 25-27.5m away, and the roadrunner lands there.
F.) In vertical, use vf^2=vi^2+2(a)(d), vf=76.7m/s.
G.) Same as B.
H.) Do the calculations with my numbers.
B.) Horizontal velocity is constant, so it will still be 3.5m/s, this is the same for G.
C.) For vertical, use d=vi(t)+.5(a)(t^2), vi=o, a=9.8, d=300m, t=?, so t=7.8s.
D.) Constant velocity horizontally, so use d=vt, so d=3.5(7.8), so d=27.3m.
E.) Yes, becasue it falls in to this, becasue it extends from 25-27.5m away, and the roadrunner lands there.
F.) In vertical, use vf^2=vi^2+2(a)(d), vf=76.7m/s.
G.) Same as B.
H.) Do the calculations with my numbers.
-
A) The components of the velocity would be horizontal at the instant he runs off the cliff and the vertical component due to the acceleration due to gravity
B) The horizontal velocity does not change with time for this problem so v =2 m/s
C) s = 1/2 at^2 s =300m, a = 9.8 m/s^2 Substitute and solve for t
D) Using t from part C, s = v x t , where s = distance and v = 3.5m/s
E) Using the distance calculated in part D, if the distance is between the value of 25m and 27.5m the road runner will get wet.
F) Using the time calculated in part C, Vf = 1/2 at^2 a= 9.8m/s^2
G) Vx = 3.5m/s for the entire problem. This assumes that no other horizontal forces are acting on the road runner such as air resistance
B) The horizontal velocity does not change with time for this problem so v =2 m/s
C) s = 1/2 at^2 s =300m, a = 9.8 m/s^2 Substitute and solve for t
D) Using t from part C, s = v x t , where s = distance and v = 3.5m/s
E) Using the distance calculated in part D, if the distance is between the value of 25m and 27.5m the road runner will get wet.
F) Using the time calculated in part C, Vf = 1/2 at^2 a= 9.8m/s^2
G) Vx = 3.5m/s for the entire problem. This assumes that no other horizontal forces are acting on the road runner such as air resistance