Question

A ball is thrown up with the velocity of 19.6 m/s
1 find the time taken to reach the maximum height
2 how high it will go

Answer 1

$\huge\mathgfrak\red{Answer}$

FIRST I MAKE SURE

• THE VELOCITY OF ANY BODY IS ZERO AT ITS HIGHEST POINT OF PROJECTILE
• THE ACCELERATION DUE TO GRAVITY IS IN OPPOSITE DIRECTION . DUE TO WHICH IT DEACCELERATE.

$initial \: velocity(u) = 19.6m {sec}^{ - 1} \\ \\ final \: velocity(v) = 0 \\ \\ acc \: to \: 3rd \: equation \: of \: motion \\ \\ {v}^{2} = {u}^{2} - 2gh \\ where \: h \: is \: max \: height \\ g = 9.8m {sec}^{ - 2} \\ \\ 0 = {(19.6)}^{2} - 2(9.8)(h) \\ \\ {(19.6)}^{2} = 19.6(h) \\ \\ h = 19.6m......(1) \\ \\ acc \: to \: 2nd \: equation \: of \: motion \\ \\ h = ut - \frac{1}{2} g {t}^{2} \\ \\ 19.6 = 19.6t - \frac{1}{2} (9.8) {t}^{2} \\ divide \: by \: 4.9 \\ \\ 4 = 4t - {t}^{2} \\ \\ {t}^{2} - 4t + 4 = 0 \\ \\ {(t - 2)}^{2} = 0 \\ \\ time \: taken \: = 2sec$