You lay on the ground shooting Velcro b***s at a velcro wall 30m away. the b***s leave the gun at the inital speed of 23m/s and you are aiming at a 35º angle. How high above the ground will the b***s stick to the wall?
+118724 Initially $$t=0$$
$$\\\ddot{x}=0 \quad \dot{x}=23cos35^0 \quad x=0\\\\
\ddot{y}=-9.8 \quad \dot{y}=23sin35^0 \quad y=0\\\\$$
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Ongoing
| $$\\\ddot{x}=0\\\\ \dot{x}=23cos35^0\\\\ x=(23cos35^0)t$$ | $$\\\ddot{y}=-9.8\\\\ \dot{y}=-9.8t+23sin35^0\\\\ y=\frac{-9.8t^2}{2}+(23sin35^0)t\\ y=-4.9t^2+(23sin35^0)t$$ |
Find y when x=30
When x=30
30=(23cos35)t
t=30/(23cos35)
t=1.592314681
$${\mathtt{\,-\,}}{\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{1.592\: \!314\: \!681}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{23}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{35}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{1.592\: \!314\: \!681}} = {\mathtt{8.582\: \!442\: \!534\: \!071\: \!419\: \!9}}$$
The ball will hit the wall 8.58 metres above the floor.
+118724 Initially $$t=0$$
$$\\\ddot{x}=0 \quad \dot{x}=23cos35^0 \quad x=0\\\\
\ddot{y}=-9.8 \quad \dot{y}=23sin35^0 \quad y=0\\\\$$
------------------------------------------------------------------
Ongoing
| $$\\\ddot{x}=0\\\\ \dot{x}=23cos35^0\\\\ x=(23cos35^0)t$$ | $$\\\ddot{y}=-9.8\\\\ \dot{y}=-9.8t+23sin35^0\\\\ y=\frac{-9.8t^2}{2}+(23sin35^0)t\\ y=-4.9t^2+(23sin35^0)t$$ |
Find y when x=30
When x=30
30=(23cos35)t
t=30/(23cos35)
t=1.592314681
$${\mathtt{\,-\,}}{\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{1.592\: \!314\: \!681}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{23}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{35}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{1.592\: \!314\: \!681}} = {\mathtt{8.582\: \!442\: \!534\: \!071\: \!419\: \!9}}$$
The ball will hit the wall 8.58 metres above the floor.
