Question

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?      

Answer 1

Hey!
given ,
u=0
t=0.5s
a)v=u+gt
v=0+10*0.5
v=5ms-1
b) average speed=total distance/total time
so,
s=ut+1/2gt^2
s=0*0.5+1/2*10^2
s=50m
so,
average speed=50/0.5=100ms-1
c) height=50m.
Hope it helps you

Answer 2

Answer:

t = 0.5 sec

a = 10m/s^2 (gravitational force)

u = 0 m/s

 

i. Using first law of motion

v = u+at

v = 0+10*0.5

v = 5 m/s

 

ii. Average speed = distance travelled/time

Using second law of motion to compute distance travelled

s = ut+1/2*a*t^2

s = 1/2*10*0.5*0.5

s = 1.25 m

 

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

 

iii. Height of the ledge would be equal to the distance travelled  

And hence, height of the ledge = 1.25m.