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

The arrow is more than 174 feet high for 2 seconds and 5 seconds.

Step-by-step explanation:

To find the time when the arrow is more than 174 feet high, we can set the equation for altitude S equal to 174 and solve for t. The equation is given by:

S = -16t^2 + v0t + s0

Substituting the given values, S = 174, v0 = 112, and s0 = 14, we have:

174 = -16t^2 + 112t + 14

Rearranging the equation and setting it equal to zero:

16t^2 - 112t + 160 = 0

Dividing both sides of the equation by 16:

t^2 - 7t + 10 = 0

Factoring the quadratic equation:

(t - 2)(t - 5) = 0

Solving for t:

t - 2 = 0 or t - 5 = 0

t = 2 or t = 5

Therefore, the arrow is more than 174 feet high for 2 seconds and 5 seconds.

Answer 2

3 seconds

Step-by-step 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* t\\\Rightarrow (-112)/(-32)=t\\\Rightarrow t=3.5 \s`

`s=ut+(1)/(2)at^2\\\Rightarrow s=112* 3.5+(1)/(2)* -32* 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=√(2as+u^2)\\\Rightarrow v=√(2* -32* 160+112^2)\\\Rightarrow v=48\ ft/s`

`v=u+at\\\Rightarrow 0=48-32t\\\Rightarrow t=(-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.