Fine the velocity

Question

The Super High Performance Stomp Rocket™ claims that it can launch a rocket straight
up into the air to a height of 400 feet. Find the total time of the rocket’s trajectory (ttotal), the initial launch speed of the rocket (v0), and its velocity 6.5 seconds (v6.5) after launch

Answer 1

h(t) = -16t² + v₀t

 

Use the kinematic relation going from 0 to 400 ft to find v₀.

 

v_f² - v₀² = 2ah_max = -2gh_max

 

v_f = 0 at h_max = 400 ft

 

v₀ = √(2gh_max) = √[2(32)(400)] ft/s = 160 ft/s

 

h(t) = -16t² + 160t = 16t(10 - t)

 

T = total trajectory time.

 

h(T) = 0 = 16T(10 - T)

 

T = 10 s

 

h(6.5) = 16(6.5)(10 - 6.5) ft = 364 ft