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Find height of a projectile

Height of a Projectile. A stone is thrown directly upward from a height of 30ft with an initial velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function
s(t)=-16t^2+60t+30. Determine the time at which the stone reaches its maximum height and find the maximum height.
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3 Answers

Finding the vertex coordinates (h, k) to get the time and the maximum height,
h = -b/(2a) = 60/(2*16) = 15/8, k = f(h) = 345/4
 
Answer: t = 15/8 sec, and the maximum height is 345/4 ft.
Hey Daniel -- here's a physical reasoning approach you may like ...
 
ave speed from start to peak is 30ft/s ... g of 32ft/s/s takes 1 7/8 sec to slow from
60ft/s to zero at peak ... ht is 30ft plus (ave speed * 1 7/8 sec) ==> 90ft less 30/8 ft
 
==> 86 1/4 ft after 1 7/8 sec ... Best wishes, sir :)
S(t) =  -16 t2 + 60t + 30
 
   This is a parabola :  Maximum( -60/ -32 , s( 15/ 8)
 
 
       t   = 15/8
 
 
 
       S( 15/8) = -16(15/8)2 +60*(15/8) + 30 = 86.25