A batter hits a fly ball that reaches a height of 40 meters. If it comes down 50 meters from home plate, how long does the fielder have to get in position to catch it?

Answer 1

The vertical component of the balls flight is completely dependent on g and the amount of time it takes to get to the highest point is the same as the time it takes to return to earth (so calculate the time to the top and double it to get time in the air). A simplified equation for this vertical motion is height = 1/2gt2, so t = square root of 40 x 2 / 10 = square root of 8 = 2.8 sec. Double that to get 5.6 sec.

Answer 2

Use the equation s=v₀sinθt -at²/2 and v= v₀ -at at 40m height v=0 so 0= v₀ sinθ -at so t=v₀ sinθ/a using this in the height equation along with the 40m =v₀²sin²θ/9.8 - 9.8v₀²sin²θ/2•9.8² which can be simplified to give 40= v₀²sin²θ/9.8(1/2) giving v₀² =80•9.8/sin²θ the range can be calculated by using x= 50 = v₀ cosθ(2t) and 25= v₀ cosθt using the equation for t gives 25=v₀²sinθcosθ/9.8. The next step is to substitute the expression for v₀² into this equation and in so doing one gets 25 =80cosθ/sinθ =80/tanθ which can be solved for θ. Doing this gives tanθ=80/25 and using the arctan (80/25) gives a θ = 72.65º. Using this in the v₀ equation gives v₀²=860.5 m²/s² and v₀= 29.3 m/s. So the time to reach 40m height is 29.3• sin72.65/9.8 and t= 2.85s the time to get ready for the ball is 2t which is 5.7s to get in position.