Question

A car falls off a ledge and drops to the ground in 0.5s.let g= 10ms-2 (what is the speed on striking the ground) ( what is its average speed during the 0.5s) (how high is the ledge from the ground)

Answer 1

Answer:

Explanation:

the car was falling from rest so

initial velocity is 0

hight

h= 0.5×10×(0.5^2)

h=10/8

h=1.25 m

and average speed =1.25/0.5 = 2.5 m/s

Answer 2

Answer:

v = 5 m/sec

s = 1.25 m

Explanation:

We can solve it easily just by using formula,

1) For Final Speed

v = u + at

here, v = Final Speed

u = Initial Speed

a = Acceleration

t = time

2) For Distance

S = ut + 1/2 at²

here, S = Total Distance covered

Units

All are in S.I unit i. e Distane in m, time is sec and Acceleration in m/sec and Speed is m/sec.

Solution :

1)

v = u + at

v = 0 + 0.5 × 10

v = 5 m/sec

2)

$\sf \: s \: = ut \: + \: \dfrac{1}{2}a {t}^{2} \\ \\ \sf = 0 \times 0.5 \: + \frac{1}{2} \times 10 \times {(0.5)}^{2} \\ \\ \sf = 5 \times 0.25 \\ \\ = \sf \: 1.25 \: m$