A particle is dropped from the top of a tower.During its motion it cover 9/25 part of height of tower in the last 1 s then find the height of tower
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let x be the height of tower.
u=0
a=+g=+9.8m/sec²
s=x
9/25x=ut+1/2at²
9/25x=0×t+1/2×9.8×1²
9/25x=4.9
x=4.9×25/9=122.5/9m=height of tower
u=0
a=+g=+9.8m/sec²
s=x
9/25x=ut+1/2at²
9/25x=0×t+1/2×9.8×1²
9/25x=4.9
x=4.9×25/9=122.5/9m=height of tower
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It's the perfect answer given by ajay..
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