A ball is thrown with a speed of 0.5m/s.(I) how high will it go before it begins to fall? (2)how long will it take to reach that height ?

Time to reach maximum hight is obtained using the formula, v² = u² -2gh ................(1)

where v is final speed which is zero at maximum height h, u is initial speed and g is acceleration due to gravity.

 

hence from eqn.(1),  h = u² /(2g)  = (0.5×0.5) / (2×9.8) = .013 m = 13 cm

 

time t to reach maximum height h is obtained using the formula v = u - g×t  or   t = u/g = 0.5/9.8 ≈ .05 s