If a boy on top of a 64 foot tower throws a ball horizontally so that the ball lands 108 feet away. How fast was the ball moving the moment it left his hand? a)54 feet/sec b)27 feet/sec c)108 feet/sec Please advise, Thank you

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If a boy on top of a 64 foot tower throws a ball horizontally so that the ball lands 108 feet away. How fast was the ball moving the moment it left his hand? a)54 feet/sec b)27 feet/sec c)108 feet/sec Please advise, Thank you

Physics
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is it 108 feet from where the ball was thrown?
|dw:1444539098522:dw| which value is 108?
okay i will assume that it is 108 feet from where the ball was thrown. (the diagonal)

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i'm thinking the 108 feet is the x measurement on your drawing.
yes it is 108ft, its the distance from where it was thrown to the stop
x is the 108ft
the first thing you need to do is find the amount of time that the ball will take to each the ground and no matter what the speed, the value will always be the same because the object will fall at a constant rate. To find this time we will need to use our first known which is the height of the tower (64 feet) we will need to convert this to meters to make it easier to solve. 64 feet = 21.333m. the next known is the rate at which it falls 9.8m/s^2. so we can use the equation s=1/2gt^2 or \[t=\sqrt{\frac{ 2s }{ g }}\] and we will get a time of \[t=\sqrt{\frac{ (2)(21.333) }{ 9.8 }}\] t=2.08s. assuming that zephyr is correct then all we need to do is find out what speed the projectile will need to be going in order to hit the ground at 108feet=36m. and speed =distance/time which means speed=36m/2.08s=17m/s. conveerting this to feet/second we get 51.92feet/second.
so if (y) was 108 feet you will need to use Pythagoras's theorem to find the bottom displacement.|dw:1444540167438:dw| \[x=\sqrt{11664-4096}\] this will be = 86.9feet=29m speed=distance/time=29/2.08=13.94m/s convert it to feet/sec and you get 41.82feet/second.
your answer is (a)
hope i helped you.
Thank you it was very well explained! I get it now!! :)
Thank you :)

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