## Physics (10th Edition)

$v^{2}=u^{2}+2ax$ $(\frac{u}{2})^{2}=u^{2}+2(-9.8)(4.00)$ $\frac{3}{4}u^{2}=78.4$ $u\approx10.22415$ m/s To find the maximum height reached: $v^{2}=u^{2}+2ax$ $x=\frac{(0)^{2}-(10.22415)^{2}}{2(-9.8)}\approx5.33$ m