## Essential University Physics: Volume 1 (4th Edition)

a) $\theta = 60.74^{\circ}$ b) $v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$
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: $v_0=\sqrt{\frac{5g}{sin^2(60.74)}}=8.03 \ m/s$