# PHYSICS: pls help me with this?

A car falls off a ledge and reaches the ground in 0.5 seconds, take g = 10m/s^2

1. What is its speed on striking the ground?

2. What is its average speed during 0.5 sec?

3. How high is the ledge from the ground?

tnx

Relevance

Write out variables:

U (initial velocity) = 0

V (final velocity) = Find out

G = 10 (acceleration)

T = 0.5

S = Displacement

v=u + at

v^2 = u^2 + 2as

s = ut + 1/2 a t^2

1)

we are going to use v = u + at

v = 0 + 10 * 0.5

v = 5ms^-1

2)

average speed is equal to distance over time

S = distance

s = ut+1/2 at^2

s = 0*0.5 (strike out) + 0.5*10*0.5^2

s= 5 * 0.5^2

s = 5*0.25

s = 1.25ms^-1

S/T = ave speed

1.25/0.5 = 2.5 ms^-1

3)

1.25 metres. We established that earlier.

Hoped i helped!! :)

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• (3) Let the height of the ledge is h meter

=>By s = ut + 1/2gt^2

=>h = 0 + 1/2 x 10 x (0.5)^2

=>h = 1.25 m

(1) By v = u + gt

=>v = 0 + 10 x 0.5

=>v = 5 m/s

(2) By average velocity (u) = h/t = 1.25/0.5 = 2.5 m/s

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• Ok, initiall assumptions:

- the car's initial velocity is: v0 = 0 (both on horizontal and vertical axis);

- there is no friction with the air during the fall;

Solution:

v = v0 + g * t;

1. Upon striking the ground: v(0.5) = 5 [m/s];

3. The motion law in accelerated motion is:

x = x0 + v0 * t + (a * t^2) / 2;

x(0.5) = 1.25 [m];

2. The average velocity is:

vav = x / t;

vav(0.5) = x(0.5) / 0.5t = 2.5 [m/s].

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• 1. striking the ground its 0 m/s

2. v = a*t + V0, v=0.5*10 = 5 m/s

3. d=v*t, d= 5*0.5= 2.5 m

hope i helped

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