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# 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?

Whoever answers first i will choose that answer as best answer. pls answer ASAP.

tnx

### 5 Answers

- 1 decade agoFavorite Answer
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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- FiremanLv 71 decade ago
(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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- 1 decade ago
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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- ?Lv 41 decade ago
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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- 1 decade ago
for part 1: t= 0.5s. a=10m/s^2

s = ut + 0.5at^2 = 0 + 0.5*10*0.5^2 = 1.25m/s

part 2: s=d/t

v = u+at = 0.5*10 = 5m/s

part 3: s = d/t --> d=ts = 0.5*5 = 2.5m

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