Lauren K.

asked • 05/16/18# If a ball is thrown into the air at an initial velocity of 48 feet per second from a height of 3 feet. After how long does the ball reach maximum height?

h(t)= -16t^2 ^ vo + ho

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## 2 Answers By Expert Tutors

Alexander B. answered • 05/16/18

Advanced Math and Science Tutoring and Test Prep online (SAT, ACT)

This math word problem has a very simple solution: t = V

_{0}/g (no need for that h(t) formula).First, convert the speed: V

_{0}= 48 ft/sec = 14.63 m/sec (approx)Divide by the value of g (approx 9.8 m/sec^2) to get the answer: t = 1.5 sec (rounded)

Note: initial height h

_{0}is irrelevant to this particular problem, unless there was an additional assignment to calculated the maximum height in addition to time tNaomi S. answered • 05/16/18

Patient and experienced math tutor

h(t)= -16t^2 +vo + ho

given that

initial velocity = 48 ft/s

initial height = 3 ft

h(t) =-16t

^{2 }+ 48t +3t =-b/2a =-48/2(-16)=48/32 =3/2 =1.5

So after 1.5 seconds ball reach to maximum height

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Thomas R.

-16t² + Vh_{0}t +_{0}) an no values for some of them. That would create a variable answer.05/16/18