Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)

Question

Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)

 

A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 69 feet per second. How high will the ball go? (Round your answer to two decimal places.) 

 

 

? ft

Answer 1

You can also solve this using the properties of a parabola.  The height h as a function of time is given by

 

h(t) = (-1/2)gt² + v₀t + h₀ = (-1/2)32t² + 69t + 6

 

h(t) = -16t² + 69t + 6,

 

with h in feet, t in sec.

 

Max h occurs at the vertex of the parabola for h(t).

 

t = -b/(2a) = -69/[2(-16)] ≅ 2.156 s

 

h_max = h(2.156) = -16(2.156)² + 69(2.156) + 6 ≅ 80.39 ft

 

Note this is the same height you get from Peter D.'s solution.

Answer 2

v² - v²₀ = 2a(y - y₀)                                 at the top, v = 0ft/sec

0 - (69ft/sec)² = 2(-32ft/sec/sec)(y - 6ft)

solve for y to get the height where v = 0