A stone is thrown vertically upward with an initial velocity of 20 metre per second taking G is equal to 10 metres per second find the maximum height reached by the stone find the net displacement and the total distance covered by the store and 4 seconds
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h max=usqr sin sqr theta/2g
=20×20×1/2×10=20m
net displace ment=0
time of flught=2×u×sin theta/g=2×20/10=4
therefore after 4 sec=distance=2×hmax=40m
=20×20×1/2×10=20m
net displace ment=0
time of flught=2×u×sin theta/g=2×20/10=4
therefore after 4 sec=distance=2×hmax=40m
rohanrit:
correct
Thnku so much
Bt what if I have to ask u some.more questions
post your questions on brainly someone would help
as this is my friends phone
i cannot answer everytime
Y dont u log in on brainly
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h=v^2/2g
h=20^2/2*10
h=20m
h=20^2/2*10
h=20m
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