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Unformatted text preview: Redirected motion
First solution: Let v be the speed right after the bounce, which is the same
as the speed right before the bounce. If t1 is the time to hit the surface, then
gt2 /2 = h − y gives t1 = 2(h − y )/g , and so v = gt1 = 2g (h − y ). The vertical
speed is zero right after the bounce, so the time it takes to hit the ground is given
by gt2 /2 = y . Hence t2 =
2y/g . The horizontal distance traveled is therefore
d = vt2 = 2 y (h − y ). Taking the derivative, we see that this function of y is
maximum at y = h/2. The corresponding value of d is dmax = h.
Second solution: Assume that the greatest distance, d0 , is obtained when y = y0 ,
and let the speed at y0 be v0 . Consider the situation where the ball falls all the
way down to y = 0 and then bounces up at an angle such that when it reaches
the height y0 , it is traveling horizontally. When it reaches the height y0 , the ball
will have speed v0 (by conservation of energy, which will be introduced in Chapter
5), so it will travel a horizontal distance d0 from this point. The total horizontal
distance traveled is therefore 2d0 . So to maximize d0 , we simply need to maximize
the horizontal distance in this new situation. From the example in Section 3.4, we
want the ball to leave the ground at a 45◦ angle. Since it leaves the ground with
speed 2gh, you can easily show that such a ball will be traveling horizontally at a
height y = h/2, and it will travel a distance 2d0 = 2h. Hence, y0 = h/2, and d0 = h.
projectile hits the plane, we have (using the above value of t)
tan θ = Chapter 4 2
gt − v cos β
v sin β
sin β cos β
sin β (50) Setting the derivative of this equal to zero to obtain the minimum θ, we ﬁnd tan β =
1/ 2 =⇒ β ≈ 35.3◦ . The associated θ is then given by tan θ = 2 2 =⇒ θ ≈ 70.5◦ . Oscillations
3.52. Increasing distance (a) Since x = (v cos θ)t and y = (v sin θ)t − gt2 /2, the square of the distance from
you is 2
= x2 + y 2 = (v cos θ t)...
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