A rock rolling horizontally at 2m/s falls off a ledge 20 meters high?

Question

a. How long did it take the rock to reach the ground?

b. Neglecting air resistance, how far did the rock travel horizontally before hitting the ground?

c. What was the rock’s vertical velocity hitting the ground?

d. What was the rock’s horizontal velocity hitting the ground?

e. What was the rock’s resultant velocity hitting the ground?

I don't want anybody to give me any of the answers. I'm really looking on how to set up the problem to get to the answer. I just started physics and I want to understand it because it's interesting but I'm having some problems. Any help is greatly appreciated. Thanks.

Answer 1

there are 2 impt eqns in dynamics.
1) v = u + at, where v is final velocity, u is original velocity, a is acceleration, and t is time.

2) s=ut + (1/2)at^2, where u, a, and t are mentioned above and s is distance moved

now to do your qn you need to resolve your problem in the horizontal direction and the vertical direction.

a) using s=ut + (1/2)at^2 in vertical direction,
20=0(t) + (1/2)(10)t^2 , assuming g(acceleration due to gravity)=10. g works only in the vertical direction. u = 0 as the original vertical velocity is 0. the 2m/s is the horizontal velocity.
5t^2=20
t^2=4
t=2 s

b) using s=ut + (1/2)at^2 in horizontal direction,
s=2(2)+0, since there is no acceleration in the horizontal direction.
s=4m

c) using v=u+at in vertical direction,
v=0+10(2)=20m/s.
remember a in vertical direction = g = 10m/s^2

d) using v=u+at in horizontal direction,
v = 2 + 0 = 2m/s as there is no acceleration in the horizontal direction.

e) since rock’s vertical velocity=20m/s and rock’s horizontal velocity=2m/s, resultant= (20^2 + 2^2)^(1/2)=20.1m/s (to 3 significant figures)

Answer 2

An object's horizonal and vertical velocities are independent. So for a), it's no different from calculating how long an object would take to fall 20 m if dropped from rest. It's y = 0.5*gt^2, where g is 9.8 m/s^2, y is the given height, and t is the unknown. For b), you just multiply the initial horizontal velocity by the time you calculated, since x = vt when you ignore air resistance. For c), use v = gt with the solution to a). d) is just the initial horizontal velocity, because the horizontal velocity doesn't change in projectile motion when we ignore air resistance. For e), you need to use the solutions to c) and d) as the legs of a right triangle, where the resultant velocity is the hypotenuse. You may want to use subscript x and y to keep the two velocities separate, as in v-sub-x and v-sub-y.