Question

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle?
Answer
224 feet
272 feet
128 feet
484 feet

Answer 1

Answer : $H = 484 feet$

Explanation :    

Given that,  

Initial height = 160 feet

Initial speed = 144 feet/ sec

Using third equation of motion

$v^{2} = u^{2} + 2gh$

$0 = (144)^{2}\ feet/s+2\times -32feet/s^{2}\times h$

$h = 324 feet$

Now the maximum height

H = h + initial height

$H = 324 feet + 160 feet$

$H = 484 feet$

Hence, this is the required solution.

Answer 2

Assuming Earth's gravity, the formula for the flight of the particle is: 

s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. 

This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. 

Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet.