The ball moves in projectile motion. When it is moving horizontally, v(y) =0
+x is directed to the right, +y is directed upward, a(x)= 0, a(y) =-g
v₀(y)=sqrt(2gh) = sqrt(2•9.8•4) = 8.85 m/s
v= v₀(y)-gt
At the top point v=0 =>
t= v₀(y)/g =8.85/9.8 =0.9 s.
The horizontal component of the ball’s velocity relative to the man is
sqrt{v₂²-v₀(y) ²} = sqrt{18²-8.85² } =15.67 m/s:
the horizontal component of the ball’s velocity relative to the hoop is
15.67 +15 =30.67 m/s,
and the man must be 30.67•0.9 = 27.6 m in front of the hoop at release.
Relative to the flat car, the ball is projected at an angle
tanα = v₀(y)/15.67 =8.85/15.67=0.565
α=29.46⁰
Relative to the ground the angle is
tanβ =8.85/(15.67 +15) =8.85/30.67= 0.289
β=16.1⁰
In both frames of reference the ball moves in a parabolic path.
The only difference between the description of the motion in the two frames is the horizontal component of the ball’s velocity.
A person is riding on a flatcar traveling at a constant speed v1= 15 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop. The hoop is at a height h=4 m above his hand. He throws the ball with a speed v2= 18 m/s with respect to the flatcar. Let g=10 m/s2 and neglect air drag completely
When the ball leaves his hand, what is the direction of the velocity vector of the ball as seen from the ground? (angle αground with respect to the horizontal in degrees
4 answers
Thanx a lot
how did you find distance x?
thank u very much elena ..that was a great help..! :)
3 answers