If t1 is the time to hit the surface then gt2 2 h y

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 1 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 2 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 cos β y ˙ gt − v cos β = = − . x ˙ 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)...
View Full Document

This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.

Ask a homework question - tutors are online