Question

12. A car falls of a ledge and drops to the ground in 0.5 s. Let g= 10 ms-2
a. What is its speed on touching the ground?
b. What is its average speed during 0.5s?
c. How high is the ledge from the ground? 3

Answer 1

Answer:

Explanation:

Car does not have vertical velocity when it starts falling from the ledge, so Vo= 0. considering downward direction as positive,

1.

$V = Vo + gt\\V = 0 + 10*0.5$

V = 5 m/s

now just before car touches the ground, its speed is 5m/s and just after touching the ground it rebounds with same speed ( considering elastic collision with ground) so,

2. average speed during 0.5 s = $\frac{5+5}{2}$ = 5 m/s

(Note: if you are asked the average velocity, it would be zero. because after touching ground the velocity becomes -5 m/s)

3.

$H = VoT + \frac{1}{2}gT^2$   putting values

$H = 0*0.5 + \frac{1}{2}*10*0.5^2$

$H=1.25$ m