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
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.
Thanx a lot
how did you find distance x?
thank u very much elena ..that was a great help..! :)
To determine the direction of the velocity vector of the ball as seen from the ground, we can break down the motion of the ball into horizontal and vertical components.
First, let's consider the horizontal component. Since the person on the flatcar is traveling at a constant speed v1 = 15 m/s with respect to the ground, the horizontal velocity component of the ball when it leaves the person's hand will also be 15 m/s.
Now, let's consider the vertical component. The ball is thrown with a speed v2 = 18 m/s with respect to the flatcar, and it is thrown upward. The initial vertical velocity component of the ball is given by v2 sin(θ), where θ is the angle between the direction of throw and the horizontal. Since we want the ball to move horizontally as it passes through the hoop, we need the vertical component of the velocity to be zero at the moment it passes through the hoop.
So, we can set v2 sin(θ) - g t = 0, where t is the time it takes for the ball to reach the hoop. Solving for θ, we have θ = arcsin(gt/v2).
Finally, we can find the angle α_ground between the velocity vector of the ball and the horizontal as seen from the ground. This angle is given by the arctangent of the vertical component of the velocity divided by the horizontal component of the velocity.
α_ground = arctan(v2 sin(θ) / v1)
Substituting the value of θ we calculated earlier, we have:
α_ground = arctan(v2 sin(arcsin(gt/v2)) / v1)
Plugging in the given values:
α_ground = arctan(18 sin(arcsin(10*4/18)) / 15)
Evaluating this expression gives the angle α_ground in degrees.