A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward. (A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s

a) 98 m/s

b) 19.6 s

c)  1449.8 m

d)  29.6 s

e)  168.6 m/s

f)  46.8 s

Explanation:

Given that

Acceleration of the rocket, a = 5 m/s²

Altitude of the rocket, s = 960 m

a)

Using the equation of motion

v² = u² + 2as, considering that the initial velocity, u is 0. Then

v² = 2as

v = √2as

v = √(2 * 5 * 960)

v = √9600

v = 98 m/s

b)

Using the equation of motion

S = ut + ½at², considering that initial velocity, u = 0. So that

S = ½at²

t² = 2s/a

t² = (2 * 960) / 5

t² = 1920 / 5

t² = 384

t = √384 = 19.6 s

c)

Using the equation of motion

v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that

0 = 98² + 2(-9.8) * s

9600 = 19.6s

s = 9600/19.6

s = 489.8 m

The maximum altitude now is

960 m + 489.8 m = 1449.8 m

d)

Using the equation of motion

v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that

0 = 98 +(-9.8 * t)

98 = 9.8t

t = 98/9.8

t = 10 s

Total time then is, 10 + 19.6 = 29.6 s

e) using the equation of motion

v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,

v² = 0 + 2 * 9.8 * 1449.8

v² = 28416.08

v = √28416.08

v = 168.6 m/s

f) using the equation of motion

S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s

1449.8 = 0 + ½ * 9.8 * t²

2899.6 = 9.8t²

t² = 2899.6/9.8

t² = 295.88

t = √295.88

t = 17.2 s

total time in air then is, 17.2 + 29.6 = 46.8 s

it's a b and

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