velocity and height

Question

If an archer shoots an arrow straight upward with an initial velocity of 128 ft/sec from a height of 6 ft, then its height above the ground in feet at time t in seconds is given by the function h(t)=-16t^2+128t+6.

 

a. What is the maximum height reached by the arrow? 

b. How long does it take for the arrow to reach the ground? 

 

(Simplify your answer)

 

thank you in advance for answering!!!

Answer 1

(a)  Maximum height occurs when speed drops to zero:

 

v(t) = v0 + at

 

a = -32 ft/s², v0 = 128 ft/s

 

v(t) = 128 - 32t = 0 ⇒ t = 4 s

 

h(4) = -16(4)² + 128(4) + 6 ft = 262 ft

 

h_max = 262 ft

 

(b)  It reaches ground when h(t) = 0 

 

h(t) = -16t² + 128t + 6 = 0 

 

t = {-128 ± sqrt[(128)² - 4(-16)(6)]} / 2(-16) = (-128 ± 129.4913) / (-32)

 

t must be positive, so use the negative square root: 

 

t = (-128 - 129.4913) / (-32) = 8.0466 s, so it reaches the ground 8.0466 s after being shot up.