A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 11.0 m/s. Assuming negligible air resistance,
1 answer:
You might be interested in
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
Answer:
(a) 0.942 m
(b) 18.84 m/s
(c) 2366.3 m/s²
(d) 0.05 s
Explanation:
(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.
(b) Its frequency, f, is 1200 rev/min = rev/s = 20 rev/s.
Its angular frequency, ω = 2πf = 2π × 20 = 40π
The speed is given by
v = ωr = 40π × 0.15 = 6π = 18.84 m/s
(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²
(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.
T = 1/20 = 0.05 s