Question

A car falls off a ledge and drops to the ground in 0.7 seconds.
Let g=10ms
−2

(a) What is its speed on striking the ground​

Answer 1

Gɪᴠᴇɴ:-

time taken=0.5s

g=10ms −2

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

(a) Under free fall,

$\sf \: u=0$

$\sf \: a=g$

$\sf \: v=u+at$

$\sf \: v=0+10×0.5$

$\sf \: v=5ms−1$

$\sf(b) Average \: speed = \frac{u + v}{2}= \frac{0 + 5}{2} =2.5ms - {}^{1}$

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

$\sf \: h=ut+ \frac{1}{2} gt {}^{2}$

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

$\sf \: h= \frac{1}{2} ×10×(0.5) 2 =5×0.25=1.25m$

Answer 2

Explanation:

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