A stone is catapulted rightward with an initial velocity of 16.0 m/s at an angle of 46.0° above level ground.?
A stone is catapulted rightward with an initial velocity of 16.0 m/s at an angle of 46.0° above level ground. Neglect air resistance. Find its horizontal and vertical displacements at the following times after launch.
(a) 0.70 s
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1. Break it up into components.
You first need to find the x and y component for v1.
V1x=(16)Cos(46)=11.1m/s
V1y=(16)Sin(46)=11.5m/s
X
v1=11.1m/s
v2=v1
t=0.70
d
a=0
*The reason why v1=v1 is because there is no air resistance. When there is no air resistance the velocity is constant. If velocity is constant then acceleration is 0.
To find the horizontal displacement use the x component values to find D.
d=v(t)
d=11.1(0.70)
d=7.77m
Y
v1=11.5m/s
v2=----
t=0.70
d=
a=-9.8
To find the vertical displacement use the y component to find D.
Since u have 3 variables u can cross out v2 since it is not needed.
Equation use.
d=v1(t)+1/2a(t)^2
d=11.5(0.70)+1/2(-9.8)(0.70)^2
d=8.05+(-2.401)
d=5.65m
Therefore the vertical distance is 5.65m and the horizontal distance is 7.77m