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 + 42t + 100 and t represents the number of seconds elapsed since the rock was thrown. How high is the bridge? feet How high above the water (in feet) is the rock 0.5 seconds after it was thrown? feet After how many seconds does the ball hit the water? seconds After how many seconds does the rock reach its maximum height above the water? seconds What is the maximum height of the rock above the water? feet

Solution

(a) The height of bridge is when t=0, that is initial height of rock

f(0) = 0+0+100 = 100ft

Hence height of bridge = 100ft

(b) f(0.5) = -16(0.5)²+42(0.5)+100

=> f(0.5) = -4 + 21 + 100 = 117

The height after 0.5 s is 117ft above the surface of water

(c) The ball hits the water, when f(t) = 0

=> -16t² + 42t + 100 = 0

=> 8t² - 21t - 50 = 0

=> t = 4.1361

Hence the ball hits the water after 4 sec

(d) f\'(t) = -32t + 42 = 0

=> -32t = -42

=> t = 1.3125

Hence the rock reaches maximum height after 1 sec above the water

(e) f(1.3) = -16(1.3)² + 42(1.3)+100 = -27.04+54.6+100 = 127.56

The maximum height is 128 ft (approx)