A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 30 m high. When it hits the ground at the base of th e cliff, the rock has a speed of 33 m/s . Assuming that air resistance can be ignored, find the initial speed of the rock.

Question

A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 30 m high. When it hits the ground at the base of th e cliff, the rock has a speed of 33 m/s .  Assuming that air resistance can be ignored, find the initial speed of the rock.

lgbasallote · Answer

you can split it up into two parts:

1) going up
2) going down

going up:
initial speed = v1
final speed = 0
acceleration = -9.8
distance = d

going down:
initial speed = 0
final speed = 33
acceleration = 9.8
distance = d + 30

so the equation for going up is $$V_2 ^ 2 = V_1 ^ 2 + 2as$$
substitute..
$$0 = V_1^2 - 19.6d$$
$$19.6d = V_1^2$$
$$d = \frac{V_1^2}{19.6}$$

now equation for going down
$$V_2 ^2 = V_1^2 + 2as$$
$$33^2 = 0 +9.8(d+30)$$
$$1089 = 9.8(\frac{V_1^2}{19.6}) + 294$$
$$795 =9.8(\frac{V_1^2}{19.6})$$
$$81.12 = \frac{V_1^2}{19.6}$$
$$1590 = V_1^2$$
$$39.87 = V_1$$

lgbasallote · Answer

did you follow that?

anonymous · Answer

yeah , thanks a lot . you sure are a help Igbasallote

lgbasallote · Answer

welcome