You are attending the State Fair, and at the midway games, you come across the one rare, crooked carnival game. In this game, balls have to be launched into holes in the wall. Placing the ball in the topmost hole wins you a giant cuddly bear, so you decide to aim for it. The wall is placed 2 m away from the launcher, and the top hole is 1.5 m above the launcher. The holes are designed in such a way that the ball will only go through if it is travelling horizontally when it reaches the hole. You can adjust the launch speed and angle of the ball. What settings for the speed and angle (measured from the horizontal) should you set the launcher in order to win the cuddly bear.

Question

Physics

Jorge Lara

27 June, 19:16

You are attending the State Fair, and at the midway games, you come across the one rare, crooked carnival game. In this game, balls have to be launched into holes in the wall. Placing the ball in the topmost hole wins you a giant cuddly bear, so you decide to aim for it. The wall is placed 2 m away from the launcher, and the top hole is 1.5 m above the launcher. The holes are designed in such a way that the ball will only go through if it is travelling horizontally when it reaches the hole. You can adjust the launch speed and angle of the ball. What settings for the speed and angle (measured from the horizontal) should you set the launcher in order to win the cuddly bear.

+3

Answers (1)

Know the Answer?

Addison White · Answer 1

At the top, horizontal displacement = 2 m.

vertical displacement = 1.5 m.

Let time to reach the hole be t. Let the ball is launched at angle θ with horizontal with velocity u.

ucosθ x t = 2

from v = u - 2gt

0 = usinθ - 2gt

usinθ = 2gt

v² = u² - 2gh

u²sin²θ = 2 g h

u²sin²θ = 2 x 9.8 x 1.5

usinθ = 5.42

usinθ = 2gt

2gt = 5.42

t =.2765 s

ucosθ x t = 2

ucosθ x. 2765 = 2

ucosθ = 7.23

usinθ = 5.42

dividing,

Tanθ =.75

θ = 37°

usin37 = 5.42

u = 9.03 m / s