Answer:
1)a. It is constant the whole time the ball is in free-fall.
2)b. = 14 m/s
3) e. = 19.6 m/s
Explanation:
1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.
2) speed = distance/time
horizontal distance = 56m
Time = 4 seconds
Speed = 56m/4s = 14m/s
3) acceleration due to gravity g = 9.8m/s^2
Initial vertical velocity = u
Final vertical velocity = v = -u
Using the law of motion;
v = u + at
a = acceleration = -g = -9.8m/s^2
t = time of flight = 4
Substituting the values;
-u = u - 4(9.8)
-2u = -4(9.8)
u = -4(9.8)/-2
u = 2(9.8) = 19.6 m/s
Initial vertical velocity = u = 19.6 m/s
Answer:
a) w = 25.1 rad/s, b) θ = 0.9599 rad
, c) α = 328.1 rad/s² d) t= 0.0765 s
Explanation: Let's work on this exercise with the equations of angular kinematics
a) The angular velocity is
w = 4.00 rev / s (2π rad / 1 rev)
w = 25.1 rad/s
b) let's reduce the angle of degrees to radians
θ = 55 ° (π rad / 180 °)
θ = 0.9599 rad
c) Let's use the initial angular velocity as the system part of the rest is zero
w² = w₀² + 2 α θ
α = w² / 2 θ
α = 25.1²/2 0.9599
α = 328.1 rad / s²
d)
w = w₀ + α t
t = w / α
t = 25.1 / 328.1
t= 0.0765 s
Answer:
1.35 miles
Explanation:
Since the slope is 14% grade = -14/100. This is a negative slope.
The distance of 1000 ft descended is a vertical descent, so assuming our initial position is 0, and Δx is the change in horizontal distance.
tanθ = -14/100 = (0 - 1000)/Δx = - 1000)/Δx
-14/100 = - 1000)/Δx
Δx = -1000 × -100/14= 7142.86 ft
5280 ft = 1 mile
7142.86 ft = 7142.86/5280 miles = 1.353 miles ≅ 1.35 miles