Answer
a) $9.96 $seconds
b) $530m$
c) $53.2\frac{m}{s}; -60.3\frac{m}{s}$
d) $80.4\frac{m}{s}$
e) $48.6^o$ below the horizon
f) $70.9 m$
Work Step by Step
a) This is in the vertical direction with up positive, down negative.
$\Delta x_f=\Delta x_i+v_{0y}t+\frac{1}{2}gt^2$
$v_{0y}=65.0\frac{m}{s}\times \sin(35.0^o)=37.3\frac{m}{s}$
$-115m=0m+37.3\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$
$-4.9t^2+37.3\frac{m}{s}+115=0$
$t=9.96s$
b) $x=v_{0x}t$
$v_{0x}=65.0\frac{m}{s}\times \cos(35.0^o)=53.2\frac{m}{s}$
$x=(53.2\frac{m}{s})(9.96s)=530$
c) $v_{fy}=v_{0y}+gt$
$v_{fy}=37.3\frac{m}{s}+(-9.8\frac{m}{s})\times9.96s=-60.3\frac{m}{s}$
The horizontal component of the velocity is constant
$v_{fy}=-53.2\frac{m}{s}$
d) $v=\sqrt{\big(53.2\frac{m}{s}\big)^2+\big(-60.3\frac{m}{s}\big)^2}=80.4{m}{s}$
e) $\arctan\Big(\frac{-60.3\frac{m}{s}}{53.2\frac{m}{s}}\Big)=48.6^o$
f) $(v_{fy})^2=(v_{0y})^2+2g\Delta y$
$(0\frac{m}{s})^2=(37.3\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$
$\Delta y_{max}=\frac{\big(37.3\frac{m}{s}\big)^2}{2\times9.8\frac{m}{s^2}}=70.9m$
