Throwing a ball, the maximum distance it will fly before hitting (level) ground

Question

Throwing a ball, the maximum distance it will fly before hitting (level) ground is achieved by throwing the ball at a 45 degree angle. The distance is then given by the equation: d = v^2 / g where v is the initial velocity and g is the acceleration due to gravity (9.8 m/sec^2). On a big, flat field, you throw a ball (at the optimal 45 degree angle), and it travels 135 feet before hitting the ground. How much harder (i.e., how much faster, as a percentage) do you have to throw the ball for it to travel 560 feet before hitting the ground?

Explanation / Answer

since we have to find how much percentage faster we have to throw the ball,

we will first subtract the two distances as speed and distance are propotional to each other and than devide it with the speed or distance from which we have to comare and will multiply it by 100 to get increase in speed or how harder we have to throw.

((560-135)/135)*100=314.81% faster

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