Explanation:
Given:
Rocket launched with
acceleration, a = +20 m/s^2 (upwards)
for
time, t = 10 seconds
coasts against gravity and eventually returns to ground.
Need
1. Maximimum height, H, reached
2. velocity on landing.
Solution:
We need to use the following kinematics equations:
v = u + at .........................(1)
v^2-u^2 = 2aS ...............(2)
where
u = initial velocity (m/s, of a particular stage)
v = final velocity (m/s)
a = acceleration (m/s^2)
S = distance travelled (m)
The rocket goes through three stages.
A. with acceleration upwards
B. no more propulsion, upwards movement (in deceleration)
C. no more propulsion, free-fall from highest point.
Stage A: with acceleration until fuel exhausted
u = 0 m/s
a = 20 m/s^2
t = 20 s
to find final velocity (at end of stage 1, apply equation (1):
v = u+at = 0 + 20*20 = 400 m/s
to find height reached, apply equation (2)
v^2-u^2 = 2aS ...............................(2) =>
S = (v^2-u^2)/2a ..............................(2a)
= (400^2-0^2) / (2*20)
= 4000 m (above ground)
Stage B: coasting upwards
u = 400 m/s (from stage A)
v = 0 (at highest point)
a = -9.81 (acceleration due to gravity, downwards, so negative)
to find distance travelled during stage B, apply equation (2a)
S = (v^2-u^2)/2a
= (0^2 - 400^2) / (2*-9.81)
= 8154.94 m
Stage C : free-fall from highest point to ground
At highest point, Total upward distance travelled (stages A & B)
= 4000 + 8154.94 = 12154.94 m (Q1)
u = 0
a = -9.81 m/s^2
S = (4000+8154.94) = 12154.94
To find velocity on impact at ground level, use equation (2)
v^2 - u^2 = 2aS ...................(2) =>
v = sqrt(2aS+u^2) ...............(2b)
= sqrt(2*20*12154.84+0^2)
= sqrt(486197.76)
= 697.3 m/s (Q2)
