Answer:
Apply the following formulae horizontally And get A value for time
Remember horizontal acceleration is zero
[tex]s = ut + \frac{1}{2}a {t}^{2} \\ 0.8 = 1.7 \times t \\ \frac{0.8}{1.7} = t \\ t = 0.47s[/tex]
and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} \times 10 \times (0.47) ^{2} \\ s = 1.1045m[/tex]
