Physics, asked by Anonymous, 1 month ago


10.2) A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s-² ( for simplifying the calculations )

i) What is its speed on striking the ground ?
ii) What is its average speed during the 0.5 s ?
iii) How high is the ledge from the ground ?

Answers

Answered by poojapugavkar
5

Answer:

Given,

time taken=0.5s

g=10ms

−2

As the car is under free fall, the initial speed is zero.

(a) Under free fall,

u=0

a=g

v=u+at

v=0+10×0.5

v=5ms

−1

(b) Average speed =

2

u+v

=

2

0+5

=2.5ms

−1

(c) From newton's equation of motion, As u=0,

h=ut+

2

1

gt

2

When h is the height, t is the time of free fall, g is the acceleration due to gravity. Putting the values in the above equation, we get

h=

2

1

×10×(0.5)

2

=5×0.25=1.25m

Explanation:

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Answered by IIMrVelvetII
31

QUESTION :- A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s-² ( for simplifying the calculations).

Given

Time taken

\sf = 0.5 \:  sec

 \sf g =  {10ms}^{ - 2}

As the car is under free fall, the initial speed is zero.

(i) What is its speed on striking the ground?

⇒ Under free fall, 

u = 0

a = g

 \fbox\green{v = u + at}

v = 0 + 10 \times 0.5

 \sf \fbox \orange{v =  {5ms}^{ - 1}}

(ii) What is its average speed during the 0.5 s?

 \sf Average \: speed =\fbox \green{\frac{u+v}{2}}

  = \frac{0 + 5}{2}

 \sf \fbox \orange{=  {2.5ms}^{ - 1}}

(iii) How high is the ledge from the ground?

⇒ From newton's equation of motion,

As  \sf u = 0,

 \sf \fbox \green{h =ut +   \frac{1}{2} {gt}^{2}}

☞ When h is the height, t is the time of free fall, g is the acceleration due to gravity.

☞ Putting the values in the above equation, we get,

 \sf h =  \frac{1}{2} \times 10 \times  {(0.5)}^{2}

 \sf = 5 \times 0.25

 \sf \fbox \orange{= 1.25m}

Hope it helps ✌

\huge\fbox \green{ \sf Thank You!!!}

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