Tricky projectile motion question, Thanks (1 Viewer)
- Thread starter shmitler
- Start date
hmmm, i tried that and still got answer wrong, these were my steps
100t-86.6t=600, therefore t=44.78
then i plugged t=44.78 into y=1/2gt^2+vsinθt +H (with v=100 and θ =30)
since we have the time of flight the value of y will be zero leading to the only unknown being H.
Thanks
What is the given answer and what answer are you getting?
hahahaha really different, answer is 88 m, im getting around -7000
For both balls we have
 &= -\frac{9.8}{2}t^2+100\sin(\pi/6)t+h \\ x_1(t) &=100\cos(\pi/6)t\end{cases})
 &= -\frac{9.8}{2}t^2+h \\ x_2(t) &=100t\end{cases})
Solve = 0)
and  = 0)
to get the time of flight for each ball (call these 
and 
).
Obviously the ball shot straight horizontally will hit the ground sooner (and will travel less), so you need to solve
-x_2(t_2) = 600)
(not the other way around!)
for
.
I calculated
.
Solve
Obviously the ball shot straight horizontally will hit the ground sooner (and will travel less), so you need to solve
for
I calculated
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