A bal is thrown vertically upwards with a speed of 19.6m/s cheejon the top of a tower returns to earth in 6 second find the height of tower
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u = 19.6 m/s t = 6s a = -10m/s^(2) s = ? v^(2) - u^(2) = 2as -(19.6)^(2) = 2 (-10)(s) -384.16 = -20 s -384.16 ÷ (-20) = s s = 19.208 m Height of the tower is 19.208 m
Answered by
1
u = 19.6 m/s t = 6s a = -10m/s^(2) s = ? v^(2) - u^(2) = 2as -(19.6)^(2) = 2 (-10)(s) -384.16 = -20 s -384.16 ÷ (-20) = s s = 19.208 m Height of the tower is 19.208 m
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