a)
using 3rd equation of motion
v^2 = u^2 + 2gh
v^2 = 5^2 + 2 * 9.8 *10
v = 14.866 m/s^2
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b)
max height it can reach
h = v^2/ 2g = 14.866^2 / 19.6 = 11.276 m
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c)
using conservation of energy
0.5 m u^2 + mgh1 = 0.5 m v^2 + mg h2
0.5* u^2 + 9.8* 10 = 0.5* 0 + 9.8* 15
u = 9.899 m/s
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