Question

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 ?
​

Answer 1

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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Answer 2

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 ✌

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