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 June 23rd, 2009, 07:59 AM #1 Newbie   Joined: Jun 2009 Posts: 2 Thanks: 0 Is this possible or impossible? Is it possible to strike a moving target twice with a box you are holding in both hands? Situation: You have a target moving left to right, 24 inches wide. You are asked to hit it twice with a 39 lb box of sand (measuring 20 inches by 20 inches), as it passes by at speeds from 1 mph to 15mph. You are holding the box with both hands and must strike it using an up and down motion. The target is approximately crotch height. Is this possible, why or why not? Create a spreadsheet showing distance covered per time and speed unit. The spreadsheet must show how fast the target has passed by at each tenth of a MPH. For example, at 0.1 mph, 24 inches passes by in how many hundredths of a second? Do the same for each 1/10th of a mph, (0.2mph, 0.3mph, 0.4mph) etc; all the way up to 15 mph. Show your formulas in your spreadsheet and explain why this is possible or impossible.
 June 24th, 2009, 11:26 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 Since 15 mph = 22 ft per second and 24 inches = 2 feet, time = distance/speed = 2ft / 0.1mph = (150/11)s = (15000/11) hundredths of a second. It would be difficult (but perhaps possible) to hit the target twice as it passes at 15mph, but the motion of the box can't be purely up/down else the target would be forced to stop when hit.
 June 24th, 2009, 11:38 PM #3 Newbie   Joined: Jun 2009 Posts: 2 Thanks: 0 Re: Is this possible or impossible? Thank-you for trying, but as expressed below, that explanation makes no sense. Please clarify. s=? The basic question really is this: "How long does it take that 2 foot space to pass by you at speeds of 0.1mph, 0.2mph, 0.3mph, etc...etc...etc......up to 15.0 mph." What is the formula for that? What is it for one foot, for 6 inches, 3 inches, 1.5 inches, .75 inches? Hopefully that will help. Thank-you again.
 June 24th, 2009, 11:50 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 The letter s denotes seconds. Do you think you need to consider the distance the target has moved from when the leading edge of the target reaches the leading edge of the box to when the trailing edge of the target passes the trailing edge of the box? That distance would be (20 + 24) inches. The formula is time = distance/speed. It would be convenient to use the distance in feet and the speed in feet per second, so that the time is given in seconds. Multiply by 100 if your time unit (for the result only) is 1/100 second. As already indicated, 15 mph = 22 ft per second, so 0.1 mph = (22/150) ft per second (or (44/25) inches per second).

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