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How far will a stone travel over level ground if it is thrown upward at an angle of 33.0 with respect to the horizontal and with a speed of 11 m per s and whats the maximum range that can be achieved?
Wilsonm516
x is horizontal, y is vertical
v=11m/s
z=angle=33°
vy=v*sin(33°)=5.99m/s
vx=v*cos(33°)=9.22m/s
we have to calculate how long the object will be in the air, the best way to calculate this is to caltulate how long does it take for the object to reach the maximum. we take it as we were to throw the object vertically up at vy speed.
v=vy + a*t v=0, because it stops at the max, so:
vy=-a*t (but im gonna leave out the minus, because we wanna know just the number)
we wanna know how much does it take for the object to reach the highest point
t1=vy/a = 0.61s (a is g=9.81m/s^2)
so tha total time of the object in the air is t=2*t1=1.22s
the vx speed does not vary with time, because there is no acceleration in the horizontal axis, so how far does the object get? v=s/t -> s=vx*t =11.25m and that is the result
LOOPDOP says: More accurately: Distance = v2 / g * sin (2 * angle)
= 112 / 9.81 * sin 66 degr. = 11.268m (rounded)
That is the distance you asked which is also the max. range.
To increase range you have to throw the stone with higher velocity than 11m/sec at 33o launch angle.
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