Answer
If $\theta=42^{\circ},$ then
$$a_{x}=45.78 \mathrm{\ m} / \mathrm{s}^{2}$$
and the final (launch) speed is
$$
v=12.54 \mathrm{\ m} / \mathrm{s}
$$
Work Step by Step
If $\theta=42^{\circ},$ then
$$a_{x}=\frac{F_{\text {net }, x}}{m}=\frac{F-m g \sin \theta}{m}$$$$=\frac{380.0 \mathrm{N}-(7.260 \mathrm{kg})\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) \sin 42.00^{\circ}}{7.260 \mathrm{kg}}=45.78 \mathrm{\ m} / \mathrm{s}^{2}$$
and the final (launch) speed is
$$
v=\sqrt{v_{0}^{2}+2 a_{x} \Delta x}$$$$=\sqrt{(2.500 \mathrm{m} / \mathrm{s})^{2}+2\left(45.78 \mathrm{m} / \mathrm{s}^{2}\right)(1.650 \mathrm{m})}=12.54 \mathrm{\ m} / \mathrm{s}
$$
