a cricket ball is dropped from a height of 20 m .find the velocity of the ball when it hits the ground and the time it takes to fall through this height
Answers
Answered by
0
from COME
1/2mV^2=mgh
=>V=(2gh)^1/2
=>V=19.79 m/s [g=9.8m/s^2]
V=U+gt
=>(v-u)/g=t
(19.79-0)/9.8=2secs
1/2mV^2=mgh
=>V=(2gh)^1/2
=>V=19.79 m/s [g=9.8m/s^2]
V=U+gt
=>(v-u)/g=t
(19.79-0)/9.8=2secs
pmtibrahim:
how did u get 19.79?
having a doubt
simply putting the values
Answered by
5
Hey!
_______________________
h=20m
u=0
v=?
t=?
g= 10 m/s
Using Newton's third law of motion, we can find the answer.
v^2-u^2=2as
v^2 - 0 = 2×10×20
v^2 = 20×20
v^2 = 400
v= √400
v= 20 m/s
Now, t=?
Using Newton's first law of motion, we can find the answer.
v-u=at
20-0=10×t
20=10t
20/10=t
t= 2s
ANSWER MAY VARY IF g= 9.8 m/s^2
_______________________
Hope it helps...!!!
_______________________
h=20m
u=0
v=?
t=?
g= 10 m/s
Using Newton's third law of motion, we can find the answer.
v^2-u^2=2as
v^2 - 0 = 2×10×20
v^2 = 20×20
v^2 = 400
v= √400
v= 20 m/s
Now, t=?
Using Newton's first law of motion, we can find the answer.
v-u=at
20-0=10×t
20=10t
20/10=t
t= 2s
ANSWER MAY VARY IF g= 9.8 m/s^2
_______________________
Hope it helps...!!!
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