Question

If an object is given an initial velocity straight upward of v 0 feet per second from a height of s 0 ​feet, then its altitude S after t seconds is given by the formula Upper S equals negative 16 t squared plus v 0 t plus s 0 . An arrow is shot upward with a velocity of 112 feet per second from an altitude of 14 feet. For how many seconds is this arrow more than 174 feet​ high?

Answer 1

Answer:

3 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

$v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow \frac{-112}{-32}=t\\\Rightarrow t=3.5 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=112\times 3.5+\frac{1}{2}\times -32\times 3.5^2\\\Rightarrow s=196\ ft$

174 feet from the ground is 174-14 = 160 ft from the launch area

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 160+112^2}\\\Rightarrow v=48\ ft/s$

$v=u+at\\\Rightarrow 0=48-32t\\\Rightarrow t=\frac{-48}{-32}=1.5\ s$

When the arrow will reach the 160 ft point while returning the initial velocity becomes equal to the final velocity which means the time taken to come down also becomes equal.

Hence, the arrow will be 1.5+1.5 = 3 seconds above a height of 174 ft from the ground.