Question

A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after blastoff, the engines suddenly fail, and the rocket begins free fall. which means that the force they produce instantly stops.(a) How high above the launch pad will the rocket eventually go?(b) Find the rocket's velocity at its highest point.(c) Find the magnitude of the rocket's acceleration at its highest point.

Answer 1

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so  its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest  point , it is undergoing free fall so its acceleration  = g