Question

A model rocket is launched with a very rapid engine burst. The engine cuts off quickly and the rocket is subject only to gravity after that (ignore air resistance).

At t=0, (which we call height=0) it is heading straight up with an initial speed of 52 m/s . We define "up" as the positive direction.

Find the rocket's altitude at 7 s?.

It says its incorrect but i got: 123.9m

52(7)-1/2 * 9.8(7)^2 = 123.9m

Answer 1

consider the motion along the vertical direction or Y-direction

V_oy = initial velocity In Y-direction = 52 m/s

a = acceleration = - 9.8

Y_o = initial position at t = 0 , = 0 m

Y_f = final position at t = 7

t = time interval = 7 sec

using the equation

Y_f = Y_o + V_oy t + (0.5) at²

Y_f = 0 + (52) (7) + (0.5) (- 9.8) (7)²

Y_f  = 123.9 m