Answer
a) $\theta = 60.74^{\circ}$
b) $v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$
Work Step by Step
We use the equation for range and for the maximum height reached to find:
$2.5 = \frac{v_0^2sin^2\theta}{2g}$
$2.8 = \frac{v_0^2sin2\theta}{g}$
Thus, we find:
$\frac{v_0^2sin^2\theta}{5g}=\frac{v_0^2sin2\theta}{2.8g}$
$.2sin^2 \theta = .357 sin\theta cos\theta$
$\theta = 60.74^{\circ}$
This means that the value of the initial speed is:
$v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$
