Answer
(a) $64.9^{\circ}$
(b) $21\frac{m}{s}$
(c) $1.8s$
Work Step by Step
(a) The launch angle can be determined as
$v_y=v_{\circ}sin\theta-gt$
This can be rearranged as:
$\theta=sin^{-1}(\frac{gt}{2v_{\circ}})$
We plug in the known values to obtain:
$\theta=sin^{-1}(\frac{(9.8)(3.97)}{2(2.15)})$
$\theta=64.9^{\circ}$
(b) We can find the initial speed as
$R=(\frac{v_{\circ}^2}{g})sin2\theta$
This simplifies to:
$v_{\circ}=\sqrt{\frac{Rg}{sin2\theta}}$
We plug in the known values to obtain:
$v_{\circ}=\sqrt{\frac{(35)(9.81)}{sin2(25.0)^{\circ}}}$
$v_{\circ}=21\frac{m}{s}$
(c) The time of flight of the bullet is given as
$t=\frac{x}{v_{\circ}cos\theta}$
We plug in the known values to obtain:
$t=\frac{35}{(21)cos25.0^{\circ}}$
$t=1.8s$
