0 like 0 dislike
14 views
A ball is thrown from $1 \mathrm{~m}$ above the ground. It is given an initial velocity of $20 \mathrm{~m} / \mathrm{s}$ At an angle of 40 degrees above the horizontal Find the maximum height reached and velocity at that point
solved

solved | 14 views

0 like 0 dislike
$\begin{gathered} x(t)=v \cos (\theta) t=20 \cos (40) t=15.3 t \\ y(t)=y_{0}+v \sin (\theta) t-\frac{9.8 t^{2}}{2} \\ y(t)=1+20 \sin (40) t-4.9 t^{2} \\ y(t)=1+12.9 t-4.9 t^{2} \\ v_{y}(t)=v \sin (\theta)-9.8 t \\ v_{y}(t)=12.9-9.8 t \\ \max \text { heightat } v_{y}(t)=0 \\ 12.9-9.8 t=0 \\ -9.8 t=-12.9 \\ t=\frac{-12.9}{-9.8}=1.3 \\ y(1.3)=1+12.9(1.3)-4.9(1.3)^{2} \\ y(1.3)=9.5 m \end{gathered}$
$y$ component of velocity is 0 at highest pt total velocity $=v_{x}=15.3 \frac{\mathrm{m}}{\mathrm{s}}$
by Platinum (122,256 points)

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike