Answer:
D because those are both concerning.
When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>. (choice-D)
Answer:
Using the range formula R = v^2 sin 2 theta / g
or v^2 = R * g / sin 86.4
v^2 = 3.14 m * 9.81 m/s2 / .998
v^2 = 30.9 m^2 / s^2
v = 5.56 m/s
This hasn't really proved the question - this would give
vy = 5.56 * sin 43.2 = 3.81 m/s
vx = 5.56 * cos 43.2. = 4.05 m/s
t = 1.57 / 4.05 = .387 sec to reach the waterfall
h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls
There seems another way to do this
vy / vx = tan 43.2 vy = .939 vx
h = vy t - 1/2 g t^2 and t = 1.57 / vx
h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2
Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s
Answer:
Distance travelled is 7 meters and the displacement is 3 meters
