Suppose a rock is thrown upward from a bridge into a river b

Suppose a rock is thrown upward from a bridge into a river below. The height of the rock above the surface of the water (measured in feet) is given by the function f where f(t) = - 16t^2 + 34t + 128 and t represents the number of seconds elapsed since the rock was thrown. a. How high is the bridge? feet b.How high above the water (in feet) is the rock 0.5 seconds after it was thrown? feet c. After how many seconds does the ball hit the water? seconds d. After how many seconds does the rock reach its maximum height above the water? seconds e. What is the maximum height of the rock above the water? feet

Solution

f(t) = -16t^2 +34t +128

a) Height of bridge ; t=0

f(0) = 128 feet

b) t =0.5 sec

f(0.5) = -16(0.5)^2 +34(0.5) +128 = 141 feet

c) falls into the water:

0 = -16t^2 +34t +128

solve the quadratic:

t = -1.95 , 4.08 . Neglect -ve root

So, t = 4.04 sec ball hits the water

d) Maxoimum height : f(t) = -16t^2 +34t +128

t = -b/2a = -(-34/(2*-16) ) = 34/32 = 1.06 sec it reaches max. height

e) Maximum height f(1.06) = -16(1.06)^2 + 34*1.06 +128

= 146.06 feet