an object takes 4 seconds in falling from the top of the tower to the ground if acceleration during fall is constant and equal to 9.8 metre per second square find the height of the Tower
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t=4sec
g/a=9.8m/s^2
h=?
and u =0
S=ut + 1/2at^2
S=1/2×9.8×4×4
s=78.4m
hope this helps you
g/a=9.8m/s^2
h=?
and u =0
S=ut + 1/2at^2
S=1/2×9.8×4×4
s=78.4m
hope this helps you
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HEY USER !!!
GIVEN::- Time = 4 secs..
Gravity= 9.8 metre..
Theirfore, height of the tower=?
Now,, From 2nd equation of motion
we have,, S= Ut +1÷2At square ..
equals to = 0×4 +1÷2×9.8×4
= 0+4.9×4 = 4.9×4×4= 78.4 Meter..
THEIFORE ,, THE REQURIED HEIGHT OF THE TOWER= 78.4 OR APPROX 79 METERR...
HOPE IT HELPS U..
FOLLOW ME FOR MORE ANSWERS..
THANKSS..
GIVEN::- Time = 4 secs..
Gravity= 9.8 metre..
Theirfore, height of the tower=?
Now,, From 2nd equation of motion
we have,, S= Ut +1÷2At square ..
equals to = 0×4 +1÷2×9.8×4
= 0+4.9×4 = 4.9×4×4= 78.4 Meter..
THEIFORE ,, THE REQURIED HEIGHT OF THE TOWER= 78.4 OR APPROX 79 METERR...
HOPE IT HELPS U..
FOLLOW ME FOR MORE ANSWERS..
THANKSS..
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