A ball is thrown up at a speed of 4.0 metre per second find the maximum height reached by the ball take g= 10 meter per second square?
Answers
Answered by
1
hey dear
here is your answer
ball will reach at the maximum height when it's final velovciuty becomese zero.
so, let we start the solution
initial velocity = 4 m/s
g = 10 m/s²
final velocity as per the above description = zero
by the third equation of motion,
v² = u²-2gs
0 = (4)²-2*10*s
16 = 20s
s = 16/20 = 4/5 = 0.8meter
so, ball will reach upto the maximum height of 0.8 meters.
hope it helps :)
here is your answer
ball will reach at the maximum height when it's final velovciuty becomese zero.
so, let we start the solution
initial velocity = 4 m/s
g = 10 m/s²
final velocity as per the above description = zero
by the third equation of motion,
v² = u²-2gs
0 = (4)²-2*10*s
16 = 20s
s = 16/20 = 4/5 = 0.8meter
so, ball will reach upto the maximum height of 0.8 meters.
hope it helps :)
drshashanka:
I think it formula is wrong ......
The right way is that
v²-u²= 2gh
no its rt. because ball is moving against the gravity so we take -g
n look ur answerr is -0.8 n as i studied height can never be negative hope u understood
Answered by
1
u=4.0 m/s
V=0m/s
g= 10m/s²
So,
v²-u²=2gh
0-16=20xh
-16=20h
h=-16/20
=-4/5=-0.8m
I hope I have answered correctly.......
V=0m/s
g= 10m/s²
So,
v²-u²=2gh
0-16=20xh
-16=20h
h=-16/20
=-4/5=-0.8m
I hope I have answered correctly.......
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