Plz help me with physics in highschool?

Question

3. a ball is rolled off a horizontal roof at 16m/s. after leaving teh roof, how long will the ball take to reach a speed of 18m/s?

7. a rock is released from the top of a 30m high cliff at the same time as a ball is thrown upwards from the base of the cliff at 20m/s. how much time elapses before they collide?

9. when a 2.0kg rock is dropped from a cliff it hits the beach at 24m/s. At what speed would a 4.0kg rock, dropped from the same cliff, hit the beach?

Answer 1

3. Horizontal speed of the ball is irrelevant

V=gt

t= V/g

t= 18.9.81=1.8 m/s

7.  

Rock falling   h1= 30 - 0.5 gt^2

Ball going up  h2= V0 t  -  0.5 gt^2

since h1=h2

solve for t

30 - 0.5 gt^2=V0 t  -  0.5 gt^2

t= 30/20= 1.5 s

9. Ignoring air resistance due to geometry of the rock the both will hit at the same time

V = gt (the mass is not in the equation)

Answer 2

If the ball rolls horizontally at 16 m/s, its initial velocity in the "y" direction is 0 m/s.  Using the kinematic equation, the time is 1.83 seconds.

Answer 3

(3)

At each instant t, the velocity of the ball is the resultant of two independent components:

- horizontal x-component: (v1) = 16 m/s (constant, neglecting the effect of air)

- vertical y-component, changing with the rate 9.81 m/s/s, starting from 0 (the initial velocity having only x-component): (v2) = gt = 9.81 t.

v^2 = (v1)^2 + (v2)^2

(v2) = sqrt[v^2 - (v1)^2] = sqrt[v^2 - 16^2]

When v = 18 m/s,

(v2) = sqrt(18^2 - 16^2) = sqrt68= 8.25 m/s

t = (v2)/9.81 = 0.84 s

The ball will reach the speed of 18 m/s after 0.84 second

(7)

Choose the ground as reference for all vertical position, upward direction as positive direction for displacement, speeds and acceleration.

We will consider 2 vertical motions under the action of gravity forces:

- for the rock:

** acceleration (pointing downward): a = -g = -9.8m/s^2

** initial velocity: 0

** initial position: +30 m (30m above the ground level)

- for the ball:

** acceleration (same): a = -g = -9.8 m/s^2

** initial velocity (upward): 20 m/s

** initial position: 0 (on ground level)

General equation for free falling motion

x = (1/2)a.t^2 + v0.t + x0

Applying to

the rock: x1(t) = -4.9 t^2 + 30

the ball : x2(t) = -4.9 t^2 +20 t

The two collide when x1 = x2, or when

-4.9 t^2 + 30 = -4.9 t^2 + 20 t

t = 30/20 = 1.5 s

(9)

The equations for distance and speed of a free falling object is

x = (1/2)at^2 + v0t + x0

v = at + v0

These equations shows that the velocity of a free falling body travelling the same distance is independent of its mass.

Therefore the 4 kg rock will hit the ground at the same speed as the 2 kg one, being released without initial velocity from the same height..