Answer
a). $t=2.55s$
b). For $t_{1}=\frac{1}{4}t$, $\theta=16.1^{\circ}$
c). For $t_{2}=\frac{1}{2}t$
$\theta=0^{\circ}$ with respect to x-direction.
For $t_{3}=\frac{3}{4}t$
$\theta=-16.1^{\circ}$ in 4th quadrant, below +x-axis
d). For $t_{1}$, Net velocity =$22.53m/s$,
For $t_{2}$, Net velocity =$21.65m/s$,
For $t_{3}$, Net velocity =$22.53m/s$,
Work Step by Step
a). $y=(u_{0}sin\theta)\,t-\frac{1}{2}gt^{2}$
$25sin30^{\circ}t-\frac{1}{2}9.8t^{2}=0$
t=2.55s
b). $t_{1}=\frac{1}{4}t=\frac{2.5}{4}=0.625s$
$v_{x}=\frac{25\sqrt 3}{2} m/s$
$v_{y}=\frac{25}{2}-\frac{12.5}{2}=\frac{12.5}{2}$
$\theta=tan^{-1}(\frac{v_{y}}{v_{x}})$
Thus, $\theta=16.1^{\circ}$
c). For $t_{2}=\frac{1}{2}t$
$v_{x}=\frac{25\sqrt 3}{2} m/s$
$v_{y}=0 m/s$
$\theta=tan^{-1}\frac{v_{y}}{v_{x}}=0^{\circ}$ with respect to x-direction.
For $t_{3}=\frac{3}{4}t$
$v_{x}=\frac{25\sqrt 3}{2} m/s$
$v_{y}=-\frac{25}{4} m/s$
$\theta=tan^{-1}\frac{v_{y}}{v_{x}}=-16.1^{\circ}$ in 4th quadrant, below +x-axis
d). For $t_{1}$, Net velocity =$\sqrt (\frac{25\sqrt 3}{2})^{2}+(\frac{12.5}{2})^{2}=22.53m/s$,
For $t_{2}$, Net velocity =$\sqrt (\frac{25\sqrt 3}{2})^{2}+(0)^{2}=21.65m/s$,
For $t_{3}$, Net velocity =$\sqrt (\frac{25\sqrt 3}{2})^{2}+(-\frac{25}{4})^{2}=22.53m/s$,
