Question
A certain freely falling object, released from rest, requires 1.40 s to travel the last 20.5 m before it hits the ground.
(a) Find the velocity of the object when it is 20.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)
m/s
(b) Find the total distance the object travels during the fall.
m
(a) Find the velocity of the object when it is 20.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)
m/s
(b) Find the total distance the object travels during the fall.
m
Answers
Henry
a. V = Vo + g*t = 0 + 9.8*1.4 = 13.72 m/s.
b. h = ho - 0.5g*t^2 = 20.5.
ho - 4.9*1.4^2 = 20.5,
ho = 20.5 + 4.9*1.4^2 = 30.1 m. = Initial ht. above gnd. = Total distance.
V^2 = Vo^2 + 2g*d = 0 + 19.6(30.1-20.5) = 188.16,
V = 13.72
b. h = ho - 0.5g*t^2 = 20.5.
ho - 4.9*1.4^2 = 20.5,
ho = 20.5 + 4.9*1.4^2 = 30.1 m. = Initial ht. above gnd. = Total distance.
V^2 = Vo^2 + 2g*d = 0 + 19.6(30.1-20.5) = 188.16,
V = 13.72
scott
acceleration during last 20.5 m ... 1.40 s * 9.81 m/s^2 = 13.7 m/s
average velocity during last 20.5 m ... 20.5 m / 1.40 s = 14.6 m/s
(velocity at 20.5 m) + (impact velocity) = 2 * average = 29.2 m/s
impact velocity = velocity at 20.5 m + 13.7 m/s
2 (velocity at 20.5 m) + 13.7 m/s = 29.2 m/s
... the velocity is negative (downward)
impact velocity / g = flight time
1/2 * g * (flight time)^2 = total fall
average velocity during last 20.5 m ... 20.5 m / 1.40 s = 14.6 m/s
(velocity at 20.5 m) + (impact velocity) = 2 * average = 29.2 m/s
impact velocity = velocity at 20.5 m + 13.7 m/s
2 (velocity at 20.5 m) + 13.7 m/s = 29.2 m/s
... the velocity is negative (downward)
impact velocity / g = flight time
1/2 * g * (flight time)^2 = total fall