A ball is projected form ground with a speed of 20 ms^(-1)

1.	A ball is projected form ground with a speed of 20 ms^(-1)
Answer» 5m 7.5m 10m 12.5m Solution :Since, `u SIN theta XX t = 10 m` `RARR 20 sin 45^(@) t = 10` `rArr t = (10)/(20 sin 45^(@)) = (1)/(sqrt(2))s` Now, `y = (20 sin 45^(@)) t - (1)/(2) g t^(2)` `= 20 xx (1)/(sqrt(2)) xx (1)/(sqrt(2)) - (1)/(2) xx 10 xx (1)/(2) = 7.5m`

Discussion