Question

A model jet is fired up in the air from a 16-foot platform with an initial upward velocity of 52 feet per second. The height of the jet above ground after t seconds is given by the equation y=-16t^+52t+16 where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot?

Answer 1

The time taken to reach the model to zero velocity
v = u + gt
0 = -52 + 9.81t
t = 52/9.81
t = 5.3 s

The maximum height of the model that can reach from the ground
y = -16t^2 + 52t + 16
y = (-16*5.3^2) + (52*5.3) + 16
y = -449.44 + 275.6 + 16
y = -157.84 feet

158 feet from the ground