Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 101: 67

$v=10.9m/s$

$x=v_{x}t=(vcos\theta )t$ $t=\frac{x}{vcos\theta}$ $h=ut+\frac{1}{2}gt^{2}=vsin\theta\frac{x}{vcos\theta}-\frac{g(\frac{x}{vcos\theta})^{2}}{2}$ Therefore, $v=\sqrt \frac{gx^{2}}{2(cos\theta)^{2}(xtan\theta-h)}$ Since, x=6.02m, $\theta=25^{\circ}$, h=1m, $v=10.9m/s$